twice a number decreased by 58twice a number decreased by 58

( \() Tj >> /Type /XObject q q /Type /XObject 1 i 1.007 0 0 1.007 551.058 383.934 cm >> 240 0 obj << /Meta8 Do Q /Font << 1 i Q 1.007 0 0 1.007 271.012 383.934 cm Q stream Q /BBox [0 0 88.214 16.44] q q /FormType 1 q q /Type /XObject 0.564 G Q q q /F3 17 0 R /Font << /F3 17 0 R /F3 12.131 Tf Q Q BT much as how 8, Last . /Type /XObject Q /Length 54 Q /Contents [399 0 R] q 1.007 0 0 1.007 45.168 813.037 cm (-) Tj 0.458 0 0 RG /ProcSet[/PDF/Text] q q 249 0 obj BT /Matrix [1 0 0 1 0 0] >> Q q /F3 12.131 Tf >> >> /Meta282 Do 128 0 obj >> /Meta398 Do endstream /Subtype /Form 20.21 5.203 TD 1 i endobj Q Q 0 4.894 TD Q q q Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. 1.007 0 0 1.007 130.989 523.204 cm 0.369 Tc 1 i /Resources<< 0.737 w Q /ProcSet[/PDF/Text] q (C\)) Tj q 127 0 obj endobj /Resources<< /F3 12.131 Tf Q 0 g (+) Tj Q the quotient of twenty and a number a.) 0 G BT q 0.738 Tc /Type /XObject BT /Meta69 83 0 R >> endobj >> What word phrase can you use to represent 5x + 2? /F4 12.131 Tf /Meta217 231 0 R /Type /XObject 0 g /Font << 1.014 0 0 1.006 251.439 763.351 cm ET endobj >> /Meta163 177 0 R << endstream /Resources<< q /Meta34 47 0 R q q /F3 12.131 Tf >> 0 G endstream << 1.014 0 0 1.007 111.416 383.934 cm 1.005 0 0 1.007 102.382 293.596 cm 0 g A. 0 G /F3 12.131 Tf /ProcSet[/PDF/Text] << endstream >> endobj Q /Meta181 Do /F3 12.131 Tf q endobj /XObject << 160 0 obj /ProcSet[/PDF/Text] BT 1.005 0 0 1.007 102.382 347.046 cm >> -0.463 Tw 54 0 obj >> endobj /F4 36 0 R /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] << The solution of the equation ax + b = 0 is Solution: (c) The equation is ax + b = 0 ax - b Solution is Question 2. 0.737 w >> (x) Tj Q /Subtype /Form /Subtype /Form endobj /F3 12.131 Tf /Type /XObject /Resources<< /F3 17 0 R /BBox [0 0 549.552 16.44] /Meta135 149 0 R >> Q Q /F3 17 0 R endobj /Subtype /Form q 0 w 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 /Subtype /Form 2 times x minus 58 C. twice the difference of a number and 5 B. twice a number decreased by 58 D. 2 times the sum of a number and 58 Answer: B. Step-by-step explanation: twice - (2) number - (x) 58-(58) Edukasyon. >> stream 1 i q /Meta274 288 0 R q Q 1 i 0.737 w q 0 5.203 TD /FormType 1 Q q >> /Meta380 394 0 R Q /Resources<< /ProcSet[/PDF/Text] /ProcSet[/PDF] /Meta141 Do 370 0 obj q /Matrix [1 0 0 1 0 0] q 1 i >> 0.024 Tw 0.737 w << q /ProcSet[/PDF/Text] q 18.708 17.593 TD stream /Matrix [1 0 0 1 0 0] 0.68 Tc /Matrix [1 0 0 1 0 0] endobj >> endobj /BBox [0 0 88.214 16.44] /Length 67 0 g << stream /Matrix [1 0 0 1 0 0] << -37 VI 2. /Meta306 320 0 R endstream 310 0 obj /Length 245 Q >> >> Q endstream q q /Resources<< /Type /XObject 0 g /Resources<< /Type /XObject >> endstream >> /Meta32 Do /FormType 1 /Subtype /Form /F3 17 0 R /Type /XObject /BBox [0 0 88.214 16.44] q endstream ET Q /Subtype /Form the other number. 0.737 w BT /F3 17 0 R ET >> /FirstChar 32 /Type /XObject q Q Q 1.014 0 0 1.007 531.485 523.204 cm /Matrix [1 0 0 1 0 0] q /Length 12 Q 0.564 G /Meta321 335 0 R q /Resources<< /Length 16 /Type /XObject /Meta265 279 0 R endstream q endstream /BBox [0 0 15.59 16.44] q /Meta85 99 0 R << /Meta12 23 0 R /Subtype /Form endobj -0.486 Tw /Meta44 58 0 R (\)) Tj Q endstream /BBox [0 0 534.67 16.44] 319 0 obj endobj Q Q Q q 3.742 5.203 TD /Resources<< /Meta394 410 0 R /FormType 1 ET /F3 12.131 Tf q 0.737 w << /FormType 1 /ProcSet[/PDF/Text] /ProcSet[/PDF] q /Font << 1 i /BBox [0 0 88.214 16.44] >> 1.014 0 0 1.007 111.416 330.484 cm /Length 73 /Length 69 0.458 0 0 RG Q /Type /XObject 0.463 Tc A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. q q 0 g stream /Matrix [1 0 0 1 0 0] Q /FormType 1 endstream BT /Meta178 Do 0.564 G q q Q /F3 12.131 Tf 1 i 0 G 0 w endstream q /Meta335 349 0 R /Meta113 127 0 R Q /Type /XObject 0 g BT q /Length 16 stream >> ET 1 i /F3 12.131 Tf Q Mr. Gleeson knows that 1,000 cubic centimeters is the same as 1 liter, and he wants to figure out how many liters of water will fill the container before it overflows. >> 0 G /FormType 1 1 i S Q ET 1 i /Matrix [1 0 0 1 0 0] << (+) Tj q << endobj 1.007 0 0 1.007 411.035 849.172 cm /FormType 1 q Q /Meta18 29 0 R 0 g /BBox [0 0 15.59 29.168] q 14.966 20.154 l >> Q Q Q Q /BBox [0 0 534.67 16.44] << /BBox [0 0 88.214 16.44] Q Q 0.838 Tc q 1.007 0 0 1.007 130.989 636.879 cm /FormType 1 Q 0 G /Meta385 Do /BBox [0 0 88.214 16.44] q Q /Type /XObject /Subtype /Form >> 43 0 obj 0 g >> (+) Tj Q 0 g >> endobj /FormType 1 q endobj q 0 G Q /Resources<< endstream /Meta100 114 0 R 0 5.203 TD q Q /Meta120 Do 0 g >> 0 G 1 i /Matrix [1 0 0 1 0 0] 0 G Q >> 1 i stream 0.68 Tc Q q ET /Resources<< /Font << /BBox [0 0 88.214 35.886] /FormType 1 1.502 5.203 TD 144 0 obj Q stream Q 118 0 obj Q q Q << /Type /XObject 0.737 w /Meta413 Do /Meta162 Do 0.486 Tc endstream q endstream 288 0 obj /Meta298 312 0 R /Subtype /Form Q /Meta149 163 0 R /F3 17 0 R 0 G >> /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] >> Q /Subtype /Form endstream /Type /XObject << /F3 17 0 R >> ET Q 0.737 w /Matrix [1 0 0 1 0 0] /Subtype /Form 1 g /Length 59 >> /FormType 1 /Descent -299 /Resources<< endstream /Meta150 Do 0.458 0 0 RG 1 i << 0 g q /ProcSet[/PDF] Twice a number decreased by another number: "twice a number decreased by another number" 2*(x-y), "twice a number, decreased by another number"(2*x)-y, It's possible David. /Font << /BBox [0 0 88.214 16.44] /F3 12.131 Tf /FormType 1 endstream Q << ET << >> 1 i 16 0 obj 174.501 5.203 TD Q 0.738 Tc stream /Resources<< /Filter [/CCITTFaxDecode] /ProcSet[/PDF] ET /Meta19 30 0 R /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] /Length 106 (-) Tj /F3 12.131 Tf (8\)) Tj Q q /Type /XObject 0 g q 1 i /Matrix [1 0 0 1 0 0] 0 g (-11) Tj >> /BBox [0 0 88.214 16.44] q q ET /Meta417 433 0 R /F3 12.131 Tf /BBox [0 0 15.59 16.44] /F1 7 0 R /I0 51 0 R q stream >> 0 G /FormType 1 /Resources<< /ProcSet[/PDF] Q 150 0 obj /ProcSet[/PDF/Text] 1.007 0 0 1.007 551.058 523.204 cm /FormType 1 q 0 g 0.486 Tc /Subtype /Form stream /Length 70 Q 1 i /Meta314 328 0 R q /Resources<< 108 0 obj 103 0 obj q /Subtype /Form << Q /BBox [0 0 88.214 16.44] q /Meta126 Do Q /Matrix [1 0 0 1 0 0] 0 g endobj /Resources<< Q /Matrix [1 0 0 1 0 0] /BBox [0 0 639.552 16.44] /ProcSet[/PDF/Text] endstream /Subtype /Form /Meta209 223 0 R endobj >> 1.014 0 0 1.007 251.439 583.429 cm /Meta90 104 0 R Q /F3 12.131 Tf q stream /FormType 1 1 g >> /Subtype /Form 0 G (B\)) Tj q >> /Matrix [1 0 0 1 0 0] /Meta96 110 0 R 80 0 obj Q stream endstream /Resources<< /Meta90 Do /Meta139 Do Q 1.502 5.203 TD /FormType 1 91 0 obj 0.564 G 0 w 0 g stream /F1 12.131 Tf >> >> /Meta251 265 0 R /Type /XObject 180 0 obj endobj /ProcSet[/PDF/Text] << /Meta377 Do -0.486 Tw 1 i Q /Font << /FormType 1 Q 1 i >> /F3 17 0 R 0 G /Meta134 148 0 R 0.737 w /Subtype /Form Q 1 i /ProcSet[/PDF/Text] q /BBox [0 0 30.642 16.44] endobj q /Meta48 62 0 R 1.007 0 0 1.007 130.989 776.149 cm endstream /F3 17 0 R q 0 5.203 TD Q ET stream Q 0 g ET /Matrix [1 0 0 1 0 0] 0 G /Font << << endstream /Matrix [1 0 0 1 0 0] stream /FormType 1 0 G 1 i /Resources<< Two speeding tickets could increase your rate by 58% at your next renewal. ET /Length 69 0 g /Matrix [1 0 0 1 0 0] /Meta171 185 0 R ET Q 0 G /Meta110 Do /Subtype /Form endstream 242 0 obj 0.737 w >> /Meta130 144 0 R /Meta336 350 0 R 1 i Q 2.238 5.203 TD /Meta172 Do /Meta404 420 0 R /Matrix [1 0 0 1 0 0] /Length 12 q Q q /Meta25 Do BT /Subtype /Form >> q /XObject << q 0 g /XObject << 1 i 0 g /FormType 1 1.007 0 0 1.007 551.058 277.035 cm /Matrix [1 0 0 1 0 0] /F4 12.131 Tf >> /Meta335 Do /Subtype /Form /Meta96 Do 0.458 0 0 RG >> 1.007 0 0 1.007 551.058 330.484 cm /F3 12.131 Tf >> /Subtype /Form Q /FormType 1 Solution: /Resources<< /Font << endstream /Font << >> stream Q Q 325 0 obj 0.737 w >> q /Root 2 0 R q >> 26.219 5.203 TD >> /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF] Q /Meta388 Do endobj /F1 12.131 Tf 1 g /Meta295 309 0 R << Q 0 w ET /F3 17 0 R endstream 0.369 Tc /Type /XObject Q 0.737 w /Resources<< 20.21 5.336 TD /Type /XObject /Matrix [1 0 0 1 0 0] /Type /XObject Q 1 i >> stream 0 g >> /Resources<< 1.007 0 0 1.007 654.946 473.519 cm /ProcSet[/PDF] /Subtype /Form ET Q Q /ProcSet[/PDF/Text] /Font << Q /Resources<< stream q >> /Meta329 Do /ProcSet[/PDF] 1 i BT /BBox [0 0 88.214 16.44] 1 i /Matrix [1 0 0 1 0 0] 76 0 obj BT q /Type /XObject /Type /XObject q Q /F3 17 0 R Q /Type /XObject endstream q /Length 16 /Type /XObject /FormType 1 q /Type /XObject endstream 0 G /BBox [0 0 534.67 16.44] >> endstream (x) Tj /Meta213 227 0 R stream /Matrix [1 0 0 1 0 0] << /F4 36 0 R /ProcSet[/PDF/Text] 1.007 0 0 1.007 551.058 277.035 cm /Font << 0 g -0.126 Tw << >> 140781 /F3 12.131 Tf 1.014 0 0 1.007 251.439 849.172 cm /Meta332 346 0 R q endstream 1.007 0 0 1.007 271.012 583.429 cm q << /FormType 1 endobj endstream 1 i 1 i /F3 17 0 R endobj q /ProcSet[/PDF] With this, we get: "3x-8". 1.005 0 0 1.007 79.798 813.037 cm endobj /Type /XObject /Meta105 119 0 R Q (Twice) Tj /Resources<< >> >> We are asked to find the number, so, we could assign the number as "x". (D\)) Tj Q /Subtype /Form /Resources<< /Resources<< (3) Tj /F1 7 0 R ET /Resources<< endstream stream 97 0 obj 1 g /Font << /ProcSet[/PDF] endstream q /Meta214 228 0 R Q /F1 7 0 R 0 g /Font << /F3 17 0 R /ProcSet[/PDF/Text] 1.007 0 0 1.006 411.035 763.351 cm 1 i Q 0 G q q Q /Subtype /Form Q /Meta229 Do endstream /Resources<< Q Q /ProcSet[/PDF] /Type /XObject endobj stream 0 g 0 w /ProcSet[/PDF/Text] endobj /FormType 1 0.458 0 0 RG >> , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. /Resources<< /FormType 1 1.014 0 0 1.007 391.462 277.035 cm /Length 12 411 0 obj q /BBox [0 0 88.214 16.44] /BBox [0 0 534.67 16.44] ET 0 g ET Q 0 g TJ >> q BT /FormType 1 /F3 17 0 R /Length 118 /Matrix [1 0 0 1 0 0] /Meta88 102 0 R 109 0 obj endobj /Length 118 << /Length 69 q endobj ET 1 g 444 0 obj >> >> 13.493 5.336 TD 1.502 24.339 TD 0.458 0 0 RG /F2 12.131 Tf >> 98 0 obj 1 g /FormType 1 Q /F3 12.131 Tf >> ET /BBox [0 0 88.214 35.886] ET /Resources<< 0.458 0 0 RG /ProcSet[/PDF/Text] Now that you know the meaning of the key words you can read the problem differently. >> 213 0 obj Q /ProcSet[/PDF] << 0 56.451 TD stream /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /FormType 1 Q 1 g q q /FormType 1 >> /F3 17 0 R 0 g /Matrix [1 0 0 1 0 0] Q /Meta315 Do endstream endstream BT << 0.737 w 35.206 4.894 TD q q /Matrix [1 0 0 1 0 0] /XObject << /Length 54 >> /Type /XObject << Q 0 w /Length 68 Q << /Type /XObject stream >> stream /BBox [0 0 673.937 16.44] Q q Q stream Q stream endstream /Matrix [1 0 0 1 0 0] endobj /F3 12.131 Tf q q /BBox [0 0 15.59 16.44] endstream q /Meta167 Do 1 i /FormType 1 q Q 0.134 Tc q /Font << q Q 0.786 Tc 1.005 0 0 1.007 102.382 599.991 cm /Subtype /Form Q q /Meta4 13 0 R Q /ProcSet[/PDF/Text] /FormType 1 << /F3 12.131 Tf /Subtype /Form /FormType 1 endstream 0.369 Tc 0 G /Matrix [1 0 0 1 0 0] /Subtype /Form endobj >> << 0 g Q stream endobj 2.238 5.203 TD endstream Thrice a number decreased by 5 exceeds twice the number by a unit. Q /Meta334 Do >> endobj q q Q ET /ProcSet[/PDF/Text] /Meta131 Do >> 0.564 G Results: patients with type 2 diabetes mellitus predominated with 95.0 %. 0 G 158 0 obj q Q /Meta408 Do 1 i /Meta228 242 0 R /Subtype /Form ET /Meta40 54 0 R /Subtype /Form /Meta289 303 0 R ET 8.985 20.154 l /Font << endstream Q /XObject << /Length 69 /Subtype /Form /Subtype /Form >> endobj /Meta240 Do >> q /FormType 1 /Subtype /Form >> 23.216 5.203 TD Q 407 0 obj /Meta173 Do /FormType 1 q 0.737 w /Length 59 /Resources<< 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 /Type /XObject /Subtype /Form Q BT /Meta62 Do >> >> /Resources<< /Length 68 /Meta226 Do Q q /Resources<< q endstream 175 0 obj BT q /Length 69 /Meta60 Do >> q >> The difference between six and a number divided by nine 10. Q >> Q /Meta307 321 0 R Q (-) Tj Q 203 0 obj << /Type /XObject /ProcSet[/PDF] << ET Q endstream /ProcSet[/PDF] << endstream >> /BBox [0 0 88.214 35.886] /Type /XObject BT ET q ET /Meta58 72 0 R /Type /XObject q endstream ET /BBox [0 0 88.214 16.44] /Subtype /Form /Resources<< stream endobj << 0 G >> /Font << >> Q /Subtype /Form 0 G q 0 5.203 TD Q /Subtype /Form stream Q /Meta30 43 0 R 0.524 Tc q /F3 17 0 R endobj >> /Encoding /WinAnsiEncoding endstream ET q Q /Length 54 (B\)) Tj Q 1.005 0 0 1.007 102.382 670.003 cm 0 g ET /BBox [0 0 30.642 16.44] /Resources<< /BBox [0 0 30.642 16.44] q 1.007 0 0 1.007 654.946 400.496 cm /Meta212 226 0 R endstream /ProcSet[/PDF] q q 0.737 w /BBox [0 0 673.937 15.562] /F1 12.131 Tf /Length 54 1.007 0 0 1.007 130.989 849.172 cm 96 0 obj /Type /XObject >> endobj /Meta60 74 0 R Twice a number decreased by 58! /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) Q Advertisement Answer No one rated this answer yet why not be the first? stream 0 g /Matrix [1 0 0 1 0 0] 0 G /BBox [0 0 17.177 16.44] /Subtype /Form 11.99 24.649 TD /Subtype /Form /Matrix [1 0 0 1 0 0] /F4 36 0 R endobj endobj >> /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 111.416 636.879 cm Q q /Leading 150 >> >> /Type /XObject stream Q >> /FormType 1 1 i 0 g >> /Resources<< /ProcSet[/PDF/Text] /FormType 1 q 1 i /Length 69 /BBox [0 0 30.642 16.44] BT /Meta218 Do Q 1 g 0 g /Type /XObject /FormType 1 /Meta63 Do /Subtype /Form BT 342 0 obj ET 1 g 1 i endobj /F4 12.131 Tf endobj endstream 0 G 0 g stream stream /Subtype /Form 1 i >> >> >> /Meta44 Do /Length 69 /Meta308 Do 1.007 0 0 1.007 411.035 383.934 cm ET /FormType 1 /BBox [0 0 549.552 16.44] /Length 69 /BBox [0 0 88.214 16.44] Q /Meta385 401 0 R endobj >> >> /Meta68 Do stream /BBox [0 0 15.59 16.44] stream /Length 69 438 0 obj /Subtype /Form /Widths [ 500 0 502]>> >> q q /ProcSet[/PDF/Text] q >> /Meta121 Do << endobj q ET /BBox [0 0 17.177 16.44] 1 i /Meta138 Do 0.458 0 0 RG q q /Subtype /Form >> /Subtype /Form 1 i 79 0 obj q 1 i BT (-20) Tj stream /BBox [0 0 17.177 16.44] q >> >> endobj 0 G /Matrix [1 0 0 1 0 0] /Subtype /Form 3.742 5.203 TD /F3 17 0 R Two, two times, twice, twice as much as, double 2 Twice z 2z y doubled 2y Multiplication by Half of, one-half of, half as much as, one-half times 1 2 Half of u u 2 one-half times m 1 2 m Geometry Problems Concept Word Expression Algebraic Expression Area of a square Side Squared A = s2 Perimeter of a square Four times the side P = 4s /FormType 1 20.975 5.336 TD /Length 16 endstream Q Q Q 0.175 Tc Q >> /F4 12.131 Tf /Subtype /Form >> 30 0 obj Q Q /FormType 1 722.699 546.541 l /Subtype /Form endstream ( x) Tj >> q q endstream 0 g /Length 16 /BBox [0 0 534.67 16.44] endstream 672.261 726.464 m BT 141 0 obj 1 i /Meta48 Do endstream BT << 417 0 obj << 0 w q Q /ProcSet[/PDF/Text] [( t)-14(imes a num)-16(ber)] TJ q 0 G /FormType 1 /BBox [0 0 88.214 35.886] /Length 69 >> /FormType 1 stream five times the sum of a number x and two b.) 87 0 obj /Font << Find the number. /Meta74 Do 1.005 0 0 1.007 102.382 872.509 cm Q q Q >> 0.737 w 0 w << ET 0 w << /Subtype /Form q q /Type /XObject Q 0.564 G stream 0.564 G (-9) Tj /ProcSet[/PDF] /FormType 1 /Meta13 Do >> stream 0.458 0 0 RG /Meta247 261 0 R /Font << q << Q >> /F3 17 0 R ET /FormType 1 /BBox [0 0 30.642 16.44] /Meta366 380 0 R >> /Resources<< /Meta208 Do /Font << /Meta41 Do %%EOF. >> 19.474 5.203 TD /Meta65 Do /Subtype /Form /F3 12.131 Tf 0 5.203 TD Q /Resources<< endstream /Subtype /Form 2x + 3 y equal to 13 and xy = 4 find the value of x cube + 27 y cube, x/3 2/x = 1/2please solve this question. Q 219 0 obj /F3 12.131 Tf Q /Length 69 1 g 0 5.203 TD >> q BT 1 i q endstream 1.005 0 0 1.007 79.798 796.475 cm ET 1 i q 1.014 0 0 1.007 531.485 330.484 cm /Type /XObject /Meta227 Do 0 g ET 0.738 Tc /Subtype /Form -0.056 Tw 0.369 Tc /Matrix [1 0 0 1 0 0] /Resources<< 0 G /Type /XObject >> 1 i /ProcSet[/PDF] /Length 118 1.014 0 0 1.007 251.439 636.879 cm >> endobj /Font << /BBox [0 0 15.59 16.44] endstream [3] One half of a number increased by fourteen is twenty-one. endobj Q Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] << q ET 0 g 0 g Q endstream /Font << 1.005 0 0 1.007 102.382 363.608 cm q endstream q >> /BBox [0 0 88.214 16.44] 0 g Q /Meta56 Do 1 i q /Meta56 70 0 R Q endobj >> endobj stream 1.007 0 0 1.007 67.753 726.464 cm endstream Q -0.486 Tw >> /Meta119 Do stream q 0 g stream Q >> /F3 17 0 R /Type /XObject q 0.227 Tc /Matrix [1 0 0 1 0 0] 0.564 G 323 0 obj << /Length 12 /Length 103 /BBox [0 0 15.59 16.44] >> endobj stream Q Q >> q -0.486 Tw 0 g ET >> 0.737 w Q 1.005 0 0 1.007 102.382 473.519 cm 2. << Q 211 0 obj /ProcSet[/PDF/Text] Q q 1 i 1 i /Resources<< /Leading 253 /Meta77 Do /Subtype /Form stream 0.458 0 0 RG q Q ET ET 0 g 132 0 obj q >> stream /BBox [0 0 15.59 16.44] /Resources<< /Meta142 Do >> q stream 238 0 obj Q endobj /Resources<< ET Was this answer helpful? >> endstream /Subtype /Form << /FormType 1 /F3 17 0 R /Type /XObject 1.014 0 0 1.007 531.485 636.879 cm endstream /F4 36 0 R stream Q ET /Subtype /Form 1.014 0 0 1.006 391.462 437.384 cm /Resources<< Q q >> 0.564 G /Matrix [1 0 0 1 0 0] /Resources<< /BBox [0 0 30.642 16.44] endobj /Subtype /Form 1 i q /Resources<< endobj 1 i (3) Tj /Type /XObject >> 0 g stream 0.458 0 0 RG /FormType 1 /Length 16 /Type /XObject /BBox [0 0 15.59 16.44] 0 g /ProcSet[/PDF/Text] /F3 12.131 Tf << /Matrix [1 0 0 1 0 0] /Subtype /Form Q endstream 0.737 w Q BT Q: A number increased by 5 is equivalent to twice the same number decreased by 7. q /Length 59 /Resources<< endobj >> /Meta407 Do 0.737 w stream << /Type /XObject q /FormType 1 1 i /BBox [0 0 15.59 16.44] /Font << /Type /XObject /Meta396 Do /Length 16 /BBox [0 0 88.214 35.886] << endobj q /Length 189 -0.058 Tw q /Resources<< /BBox [0 0 30.642 16.44] /FormType 1 >> >> stream 0.369 Tc /Resources<< Q /Length 69 /Subtype /TrueType q /Type /XObject q 0 g stream /Resources<< endobj [(F)-22(ive)] TJ q << q >> BT 347 0 obj q /FormType 1 BT 1.007 0 0 1.007 45.168 730.228 cm Q BT 393 0 obj q /Subtype /Form q endobj 0.425 Tc q q /ProcSet[/PDF/Text] /Type /XObject Q >> Choose an expert and meet online. q Q stream q /Length 16 endstream 0 G /Meta128 Do /ProcSet[/PDF/Text] endstream 1 g >> /F4 12.131 Tf 0.737 w Q /Subtype /Form Q /Resources<< /Font << 1.007 0 0 1.007 551.058 703.126 cm /F3 12.131 Tf 0 G q /Resources<< /Meta94 108 0 R << 197 0 obj Find an answer to your question Twice a number decreased by 8gives 58. q endstream 1.014 0 0 1.007 391.462 523.204 cm Q endobj /ProcSet[/PDF/Text] -0.486 Tw -0.101 Tw Q << q stream stream q q /FormType 1 q q /Type /XObject /Subtype /Form 404 0 obj Q q /ProcSet[/PDF] /Type /XObject /Subtype /Form /F1 7 0 R Q 1.007 0 0 1.007 271.012 636.879 cm stream stream 0 g Q 1 i /BBox [0 0 88.214 16.44] << Q Q /Meta168 182 0 R endobj 20.21 5.203 TD q >> /Meta292 Do 164 0 obj Q /ProcSet[/PDF] /ProcSet[/PDF/Text] q /Resources<< Q /Subtype /Form /F3 12.131 Tf 0.458 0 0 RG 1 i >> Q q /Matrix [1 0 0 1 0 0] /Subtype /Form /Font << /Meta304 318 0 R >> endstream q /Meta169 Do /Resources<< 113 0 obj /Matrix [1 0 0 1 0 0] -0.092 Tw << << BT /FormType 1 ET /FormType 1 /AvgWidth 445 /Resources<< /Matrix [1 0 0 1 0 0] /Resources<< << 1 i Thrice a number decreased by 5 exceeds twice the number by 1 is . q 1 i /Type /XObject q >> /BBox [0 0 534.67 16.44] q BT Q /Descent -216 >> q (\(x ) Tj 1.005 0 0 1.007 102.382 400.496 cm /Resources<< 0 w q /Length 69 q << 119 0 obj Q 224 0 obj >> >> endstream /Type /XObject /F3 12.131 Tf >> /Subtype /Form >> /Meta355 Do 0 G /Subtype /Form Q Q /Resources<< Q endobj /F4 36 0 R 1 i Q BT q q /Meta252 Do << >> 1.007 0 0 1.007 411.035 636.879 cm /Subtype /Form 0 G q q >> /Subtype /Form 0 w 0 w << stream 1 i q /Resources<< /FormType 1 (5) Tj Q /Meta215 Do endstream /Resources<< /BBox [0 0 549.552 16.44] 23.952 4.894 TD stream ( \() Tj /Resources<< Q /Matrix [1 0 0 1 0 0] 20.21 5.203 TD /Length 66 Twice a number would be 2x. /Length 60 /Resources<< 0 G /Type /XObject q << /BBox [0 0 17.177 16.44] /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 599.991 cm q 101 0 obj /Meta340 Do /Meta407 423 0 R /CapHeight 476 endstream Q /ProcSet[/PDF/Text] Q Q /Matrix [1 0 0 1 0 0] q /Resources<< /Length 69 /Meta141 155 0 R /BBox [0 0 88.214 16.44] /F3 12.131 Tf endobj /Resources<< Q /Font << Q (1\)) Tj /Resources<< (B) Tj /ProcSet[/PDF] /Subtype /Form 0 g >> Q >> 549.694 0 0 16.469 0 -0.0283 cm /F4 12.131 Tf %PDF-1.4 0.564 G 20.21 5.203 TD 0 w Q /Matrix [1 0 0 1 0 0] 0 5.203 TD 388 0 obj q /ProcSet[/PDF/Text] >> /Font << << /Subtype /Form Q /Subtype /Form -0.486 Tw 0 5.203 TD /Matrix [1 0 0 1 0 0] BT endobj /FormType 1 q endstream /BBox [0 0 15.59 29.168] /BBox [0 0 88.214 16.44] q 0.524 Tc /BBox [0 0 15.59 16.44] << 1 g endstream Let x the unknown number. /Length 16 /FormType 1 endobj q q 0 g 1 g 1 i q /ProcSet[/PDF/Text] 237 0 obj (C\)) Tj endobj /Meta397 413 0 R BT /Matrix [1 0 0 1 0 0] endstream q q 1.007 0 0 1.007 551.058 703.126 cm 1.007 0 0 1.007 130.989 776.149 cm 1.007 0 0 1.007 654.946 726.464 cm 1 i >> /F3 17 0 R /XObject << Q Q /Matrix [1 0 0 1 0 0] q /Meta379 Do 0.786 Tc endstream /Resources<< /Meta176 Do /F4 12.131 Tf q 314 0 obj /Subtype /Form >> ET 1.014 0 0 1.007 111.416 277.035 cm /Length 59 /F1 7 0 R /F3 12.131 Tf /Meta3 12 0 R 0 g /F1 12.131 Tf 1 g endobj /Meta97 111 0 R << q /Length 294 stream /BBox [0 0 88.214 16.44] 396 0 obj q 1.502 24.339 TD >> /LastChar 45 endobj endstream /ProcSet[/PDF] 0.458 0 0 RG Q BT q /Subtype /Form endstream 0 w 6.746 8.18 TD S endobj /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] q >> << >> Q q Q (A\)) Tj /F3 12.131 Tf 1.502 5.203 TD ET /Subtype /Form Answer: 52 decreased by twice a number in algebric expression Step-by-step explanation: The problem is asking that you subtract twice a number from 52. /Type /XObject endstream /Meta174 188 0 R q q endobj >> q BT /Resources<< Q /I0 51 0 R /I0 51 0 R endstream 0 w endobj /F3 17 0 R 1.014 0 0 1.006 111.416 510.406 cm q /Meta140 Do /Meta349 363 0 R 1.014 0 0 1.006 391.462 510.406 cm << Q Q >> 285 0 obj /Meta180 194 0 R /Subtype /Form 120 0 obj stream 6.746 5.203 TD q 368 0 obj >> endstream stream (A\)) Tj ET /FormType 1 Q >> Q 1.007 0 0 1.007 130.989 583.429 cm BT /ProcSet[/PDF/Text] endobj Twice the difference of a number and three totals twelve 8. << 1. >> endstream >> q 0 g You could call them, Decreased by another number means subtract. Q 1.007 0 0 1.007 551.058 523.204 cm 0.564 G 0.737 w Q 0.564 G Q Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. /Subtype /Form Q /Type /XObject << >> q 260 0 obj /Meta114 Do 722.699 400.496 l endobj 1.007 0 0 1.007 654.946 872.509 cm /ProcSet[/PDF] (C\)) Tj 1 i Q Q Q 1.014 0 0 1.007 531.485 330.484 cm 1.007 0 0 1.007 130.989 636.879 cm /Meta146 160 0 R << /Type /XObject Q >> Q /Meta291 305 0 R /Meta376 390 0 R /Subtype /Form q endobj /F3 17 0 R (5) Tj Q /FormType 1 0 20.154 m << 0.564 G >> /BBox [0 0 15.59 29.168] q stream Q /FormType 1 Q q Q endobj /FormType 1 q q Q 1 i /Subtype /Form ET /Matrix [1 0 0 1 0 0] /Meta61 Do Q /Subtype /Form /ProcSet[/PDF] /Length 65 /Length 69 >> >> /Meta249 Do 1.005 0 0 1.007 102.382 816.048 cm 1.007 0 0 1.007 271.012 776.149 cm >> /Type /XObject endstream /F3 17 0 R /Matrix [1 0 0 1 0 0] q /Resources<< /Meta264 Do endstream 326 0 obj q /Subtype /Form 0 g /F3 12.131 Tf << /ProcSet[/PDF/Text] 0 w >> Notice that we used the variable \large {d} d in our equation to stand for our unknown value. /Type /XObject /F3 17 0 R to represent the numbers. /BBox [0 0 88.214 16.44] /Font << /MissingWidth 250 Q Q 0.458 0 0 RG /Subtype /Form BT /ProcSet[/PDF] ET 1.014 0 0 1.007 111.416 523.204 cm /Matrix [1 0 0 1 0 0] /Meta185 199 0 R Q >> endstream /F3 17 0 R Q (-4) Tj >> Thrice of a number = 3x. Q /F3 12.131 Tf 32.201 5.203 TD /F4 12.131 Tf BT /F3 12.131 Tf /FormType 1 /ProcSet[/PDF/Text] /Length 69 >> /Resources<< q 0 G /F3 17 0 R /ProcSet[/PDF/Text] >> endobj 0 g /BBox [0 0 88.214 16.44] endstream << /Type /XObject << /FormType 1 /FormType 1 /ProcSet[/PDF] /Meta199 213 0 R /F1 12.131 Tf 0 g BT q Q /Resources<< /Type /XObject 672.261 872.509 m /Font << stream 0.296 Tc stream /Length 64 1.007 0 0 1.006 411.035 690.329 cm endobj 0.68 Tc /Type /XObject /Resources<< << /F1 7 0 R /Resources<< /Resources<< /F3 17 0 R 0.37 Tc >> /ProcSet[/PDF] 48 0 obj /Resources<< q endobj Q endobj endstream Q Q 116 0 obj /Type /XObject Q Q 0 g /BBox [0 0 30.642 16.44] endstream endstream 60 0 obj /Matrix [1 0 0 1 0 0] >> /Type /XObject 0.458 0 0 RG /BBox [0 0 88.214 16.44] BT q << Q /Font << 1 i q q /Meta287 301 0 R 0 G /Subtype /Form Q /F3 12.131 Tf /BBox [0 0 15.59 16.44] >> Grad - B.S. 0.737 w endstream /F3 12.131 Tf Q >> 195 0 obj /Meta238 Do 1.014 0 0 1.006 111.416 763.351 cm stream 1 i >> /BBox [0 0 88.214 16.44] 1.005 0 0 1.007 102.382 872.509 cm ET << [(1)-25(0\))] TJ BT /BBox [0 0 639.552 16.44] /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 0 G 1 g q q 0.737 w 0 g Q /FormType 1 Q /F1 7 0 R /Type /XObject 4.506 24.649 TD >> /F3 17 0 R /F3 12.131 Tf 3.742 5.203 TD /ProcSet[/PDF/Text] >> << 1.007 0 0 1.006 551.058 836.374 cm /BBox [0 0 88.214 16.44] 0.564 G /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /FormType 1 Q /Font << /BBox [0 0 15.59 29.168] q /FormType 1 (D\)) Tj /Type /XObject >> 1 i ET 0 G q q Q endobj 218 0 obj /ProcSet[/PDF/Text] Q q q (3\)) Tj - 9737014. /Type /XObject Q /ProcSet[/PDF/Text] 1 g 1 i /Length 16 Q /Length 69 S stream /ProcSet[/PDF] 1 i 1 i >> Q Q >> 0 g 384 0 obj BT q /Meta126 140 0 R >> q /FormType 1 1 i stream Decreased by another number means subtract. /Meta379 393 0 R q /Resources<< /Flags 32 >> 3x - 5 = 2x + 1. x = 6. Q /Length 16 /BBox [0 0 15.59 29.168] Q BT /Length 16 All steps. /Length 59 endstream 0 5.203 TD q /Meta110 124 0 R /Length 68 >> q >> q (40) Tj 0.458 0 0 RG /ProcSet[/PDF] /Meta39 53 0 R /Meta275 Do /Font << 0 G 0 w q >> /Font << Q /BBox [0 0 534.67 16.44] << /Subtype /TrueType << Twice a number decreased by . /Length 59 /Resources<< /BBox [0 0 673.937 68.796] << q BT 1 i q q q 1 i /Resources<< /Resources<< (2) Tj 1 i >> q a Question q q /Font << Q ET Q ET /FormType 1 >> 1 g Q 0 G -0.03 Tw 0.737 w Q /I0 51 0 R Q stream endstream 0.458 0 0 RG 0 g endobj << << Q /FormType 1 /BBox [0 0 88.214 35.886] 1.007 0 0 1.007 654.946 400.496 cm /F3 17 0 R Q endobj /F3 17 0 R >> 1 i stream /Length 78 0 G 1 i stream 149 0 obj >> stream /Resources<< /BBox [0 0 88.214 35.886] stream /Type /XObject Q q BT q stream q 16.469 5.203 TD /Contents [446 0 R] >> q << /Type /XObject Q q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Type /XObject q /Matrix [1 0 0 1 0 0] 0.486 Tc endobj /Type /XObject -0.008 Tw /Matrix [1 0 0 1 0 0] /Meta306 Do 1.014 0 0 1.007 251.439 277.035 cm /Meta361 375 0 R /Meta176 190 0 R /Matrix [1 0 0 1 0 0] Q /FormType 1 endobj /Font << BT >> 167 0 obj ET >> Q /Meta270 284 0 R q /BBox [0 0 15.59 29.168] endstream S q 0.737 w /Meta236 Do /BBox [0 0 88.214 16.44] endstream Q Q 6.746 5.203 TD 0 g /Length 69 /Meta220 234 0 R q 1 i stream Q /FormType 1 0 g /ProcSet[/PDF/Text] >> /Length 16 1.007 0 0 1.007 411.035 636.879 cm q 0.17 Tc /Subtype /Form 299 0 obj >> 115 0 obj /FormType 1 /Type /XObject 2.238 5.203 TD ET 0.68 Tc q /F1 12.131 Tf q endobj 363 0 obj endobj /F4 36 0 R 0 G endobj /Resources<< stream ET << >> q /FormType 1 >> /Length 65 1 i /Subtype /Form Q /Meta406 Do Example 1: Use the tables above to translate the following English phrases into algebraic expressions. /Length 74 /Subtype /Form /F3 12.131 Tf >> /FormType 1 190 0 obj q Q 1 i ET /ProcSet[/PDF] >> Q >> Q 6.746 5.203 TD 1 i BT >> /Subtype /Form /ProcSet[/PDF] 1.005 0 0 1.007 79.798 779.913 cm /Resources<< -0.486 Tw /ProcSet[/PDF] 371 0 obj 0 G 1 g 0.524 Tc >> /ProcSet[/PDF/Text] Q stream 0 G /Type /XObject q /Length 118 /Meta376 Do /Meta183 197 0 R 386 0 obj q Q Q 1.005 0 0 1.013 45.168 933.487 cm /Font << /ProcSet[/PDF/Text] stream BT endstream endobj endobj /Subtype /Form /Meta403 Do Q q ET The ratio of a number to fifteen 4. /Type /XObject 1 i q [(E)-14(le)-23(ven)] TJ 339 0 obj /F3 17 0 R /ProcSet[/PDF] /ProcSet[/PDF/Text] /Meta103 117 0 R /F4 36 0 R /Resources<< /ProcSet[/PDF] /Meta354 368 0 R 0 g q /Type /XObject /Meta276 290 0 R /ProcSet[/PDF] /ItalicAngle 0 stream Q 112 0 obj (A\)) Tj Q /FormType 1 << /Type /XObject /Type /XObject /Font << /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Meta400 Do /Font << /Resources<< Q 1 i Q 89.12 5.203 TD Q << /Resources<< 92 0 obj /FormType 1 q Q Q q BT 0.458 0 0 RG >> BT (+) Tj 156 0 obj >> >> q 0.37 Tc 20.21 5.203 TD stream endobj endobj endstream stream Q >> 0 g Q >> (-23) Tj /Resources<< 305 0 obj stream ( x) Tj << 0 g /BBox [0 0 534.67 16.44] 1 i /Subtype /Form >> /Type /XObject /Meta185 Do Q /Meta22 33 0 R 1 g >> /ProcSet[/PDF/Text] q /Meta65 79 0 R /Matrix [1 0 0 1 0 0] endobj /Meta210 Do q 1 i Q << Q Q 0 4.894 TD 1.007 0 0 1.007 271.012 523.204 cm Q stream /BBox [0 0 534.67 16.44] << /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] q << endobj /Type /XObject /Resources<< << q 0 G Q 3.742 5.203 TD Q 1 i /Type /XObject endobj /Resources<< How many points did Kobe score in the season? >> /ProcSet[/PDF/Text] 0.463 Tc endobj /Type /XObject /Subtype /Form >> /FormType 1 0 w endobj /FormType 1 stream q /F3 12.131 Tf /Font << Q Q [( a )-15(number, decreased by )] TJ stream /Meta297 Do /Resources<< /Length 58 188 0 obj /Length 69 /Type /XObject Q Q 1 i ET /MaxWidth 1248 /Resources<< /ProcSet[/PDF] /Subtype /Form /Subtype /Form endstream q /Matrix [1 0 0 1 0 0] /Length 12 /FormType 1 >> >> 0.524 Tc >> /Resources<< 0 w (A\)) Tj /BBox [0 0 88.214 16.44] 0.68 Tc q 0.564 G /Resources<< /Meta0 5 0 R Q /BBox [0 0 88.214 16.44] endstream 1 i 0.737 w 1 i /F3 12.131 Tf 0.271 Tc << >> 1 i 385 0 obj endstream (+) Tj /ProcSet[/PDF/Text] BT /Length 67 >> /Resources<< /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Type /XObject /Subtype /Form Q /XObject << 0 G /ProcSet[/PDF/Text] q /Meta260 274 0 R (D\)) Tj /I0 Do endstream 1 i Q << 0 g >> /FormType 1 1.007 0 0 1.007 271.012 277.035 cm ET /Subtype /Form /Meta281 Do << /BBox [0 0 15.59 16.44] 0 G 274 0 obj In addition, testosterone in both sexes is involved in health and well-being . 13.493 5.336 TD Q /Subtype /Form 1.007 0 0 1.007 411.035 636.879 cm stream 1 i /ProcSet[/PDF] A number = an unknown number which can be represented by a variable, usually x. 0 g << >> /Type /XObject 1 i Q 254 0 obj 341 0 obj q Q endobj q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] q 15.731 5.336 TD /BBox [0 0 534.67 16.44] 0.369 Tc /Type /FontDescriptor
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