0.425 Tc endobj Q Q If LtitnS6S . Q >> /Resources<< 0 G Q ET /FormType 1 24 0 obj /BBox [0 0 30.642 16.44] BT 0 G 0.51 Tc >> q 0.737 w Q (x) Tj 165 0 obj /Length 69 253 0 obj Q >> 0 g >> /F3 12.131 Tf /Matrix [1 0 0 1 0 0] << stream Q /Length 16 Q << >> q 295.086 4.894 TD endobj BT /F4 12.131 Tf /ProcSet[/PDF/Text] endstream Q /ProcSet[/PDF] /Resources<< stream BT endobj BT 277 0 obj q >> /F3 12.131 Tf 1 i q 0 G 0 G q 1 i >> /Resources<< /Matrix [1 0 0 1 0 0] Q >> 1.005 0 0 1.007 102.382 473.519 cm /Meta85 Do 0 G >> /Type /XObject q >> << /Subtype /Form stream q 0 G /Resources<< >> Q /BaseFont /TestGen-Regular Q 57.656 5.203 TD q >> 0 447 1 i 0 w 349 0 obj /ProcSet[/PDF/Text] >> << q /BBox [0 0 15.59 29.168] 0.737 w BT /Meta51 65 0 R /Flags 32 3.742 5.203 TD [(A numb)-16(er subtract)-15(ed from )] TJ /Type /XObject q q q /Type /XObject >> endobj q 0.737 w stream BT q q stream q /FormType 1 164 0 obj endobj >> q /BBox [0 0 88.214 16.44] (x) Tj 1.014 0 0 1.007 111.416 383.934 cm ET Q 0.737 w /Meta322 336 0 R 1 i << >> 0.564 G q (-) Tj >> Q /Length 12 Q /Subtype /Form >> 1.014 0 0 1.007 111.416 849.172 cm BT BT /Resources<< 20.21 5.203 TD endobj /Matrix [1 0 0 1 0 0] /Meta284 298 0 R /Meta375 389 0 R << q /Type /XObject Q /ProcSet[/PDF/Text] 244 0 obj /FirstChar 32 0.458 0 0 RG 0 w /Resources<< 1 i 0 g stream /Length 58 /Length 118 Q 1 i << 0.564 G Q 0 g /Encoding /WinAnsiEncoding 0 w endobj /Type /XObject 0 g ET 1.007 0 0 1.007 271.012 523.204 cm /Matrix [1 0 0 1 0 0] 0.564 G Q /Meta224 238 0 R /BBox [0 0 88.214 16.44] /Subtype /Form /Subtype /Form /Font << q ET /F1 12.131 Tf 0.425 Tc 1.007 0 0 1.007 411.035 849.172 cm >> BT endstream endobj /Length 59 q 1.007 0 0 1.007 654.946 872.509 cm /FormType 1 /Meta379 393 0 R /F3 17 0 R Q 0 5.203 TD ET 1 i 0.564 G /Meta78 92 0 R /Meta263 Do /FormType 1 /Matrix [1 0 0 1 0 0] (C\)) Tj /ProcSet[/PDF/Text] 1.007 0 0 1.007 271.012 450.181 cm /ProcSet[/PDF] Q /Matrix [1 0 0 1 0 0] /Type /XObject q /Resources<< >> /Matrix [1 0 0 1 0 0] /Length 12 q endobj /Length 59 << q ET 0.369 Tc 0.564 G 0.458 0 0 RG Q /Matrix [1 0 0 1 0 0] /Font << 1.007 0 0 1.006 411.035 763.351 cm /F3 12.131 Tf 1 g 1 g ET /Subtype /Form /Meta225 239 0 R >> endstream /ProcSet[/PDF/Text] (5\)) Tj /BBox [0 0 88.214 35.886] << endobj 1.007 0 0 1.007 271.012 703.126 cm Q endobj /BBox [0 0 88.214 16.44] /Length 59 0.31 Tc << /Length 16 (11) Tj 1 g /Length 69 0 5.203 TD /Length 118 ET /Length 60 /Subtype /Form /BBox [0 0 88.214 16.44] /Type /XObject /Resources<< /MissingWidth 252 stream /FontBBox [-170 -292 1419 1050] << >> 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 0 g /Subtype /Form 0 g /Matrix [1 0 0 1 0 0] /Type /XObject q /ProcSet[/PDF] >> (5) Tj q >> /Font << 1 g 1.007 0 0 1.007 411.035 330.484 cm /Matrix [1 0 0 1 0 0] endobj /Subtype /Form << Q endstream q 0 g q /Subtype /Form /ProcSet[/PDF] 0.425 Tc /Meta292 306 0 R q /F3 17 0 R BT View the full answer. q >> /ProcSet[/PDF/Text] ET /Font << << 1.005 0 0 1.007 102.382 546.541 cm q q 119 0 obj 1.007 0 0 1.007 411.035 330.484 cm BT /Matrix [1 0 0 1 0 0] stream 0.458 0 0 RG >> endstream /Resources<< /ProcSet[/PDF/Text] Q 1 i endobj q 1 i 0 g BT Q BT q endstream 0.425 Tc q 1.005 0 0 1.015 45.168 53.449 cm Q q 1.005 0 0 1.006 45.168 879.284 cm /Subtype /Form 30.699 5.203 TD >> /ProcSet[/PDF/Text] 0 G >> >> Q >> /Resources<< /Meta18 Do BT BT 1 g endstream /Meta209 Do /ProcSet[/PDF] >> /BBox [0 0 88.214 16.44] endobj /Resources<< >> 0 G /Type /XObject (-20) Tj 372 0 obj q endstream 0 g /Length 16 Q >> stream q /F3 12.131 Tf 0.564 G /F3 12.131 Tf /ProcSet[/PDF] BT /BBox [0 0 88.214 16.44] /ProcSet[/PDF] /Font << (B\)) Tj 1.007 0 0 1.007 271.012 330.484 cm /Meta41 Do /Meta279 Do ( x) Tj /Font << q q /Meta349 Do >> endstream q 364 0 obj Q endobj q /F1 12.131 Tf << /Meta169 183 0 R Q Q /Meta292 Do 159) n decreased by 28 is equal to 48 160) the difference of m and 27 is 34 161) a number decreased by 9 is 23 162) 13 less than w is equal to 35 163) the difference of a number and 22 is equal to 34 164) a number decreased by 27 is equal to 29 165) the difference of r and 20 is 37-6-You may use this math worksheet as long as you help someone . 0 w /F3 12.131 Tf >> 1 g q BT Q endstream /Meta117 131 0 R >> (iii) 25 exceeds a number by 7. << << -y. 229 0 obj q endstream /Meta409 Do >> endobj /Type /XObject BT /FormType 1 /FormType 1 1.007 0 0 1.007 130.989 636.879 cm q q /F3 17 0 R /Meta364 Do 404 0 obj endobj /Length 16 /Matrix [1 0 0 1 0 0] 1 i Step 1/1. Q q /Meta305 319 0 R 0 g Q /Type /XObject q Q /Resources<< /Type /XObject 0 G /Length 69 endobj q 1.014 0 0 1.006 531.485 836.374 cm /BBox [0 0 673.937 16.44] 20.21 5.203 TD /Subtype /Form (v) 5 subtracted from thrice a number is 16. /Matrix [1 0 0 1 0 0] /Subtype /Form >> /Meta17 Do /ProcSet[/PDF] /Meta286 Do endobj Check out a sample Q&A here. A) 5 more than a number. /Meta21 Do /Subtype /Form q 0 20.154 m 0 w BT Twice the number means = 2x Twice the number increase by 8 means =2x+8 Twice the number increase by 8 is 20 then means 2x+8=20 Therefore the solution to this equation will be as follows: 2x=20-8 2x=12 Divide both sides by the coefficient of. q >> 0 w >> >> /Length 294 1 g /FormType 1 /F3 17 0 R /Matrix [1 0 0 1 0 0] endstream /Meta178 192 0 R >> >> endobj Q /Length 12 /F3 12.131 Tf /F3 12.131 Tf /Resources<< endstream On This Page The Division of Cancer Prevention furthers the mission of the National Cancer Institute by leading, supporting, and promoting rigorous, innovative research and traini 0 g 0.737 w 0.838 Tc The sum of a number and 2 is 6 less than twice that number. endstream stream /F3 12.131 Tf q Q /Meta270 284 0 R >> /Matrix [1 0 0 1 0 0] /Font << endobj /Subtype /Form q /Meta222 Do /Resources<< Q (x) Tj 6.746 8.18 TD /Subtype /Form first we change the sentence in formula as the following, i think this is clear &easy to understand .i hope it helps you, This site is using cookies under cookie policy . /ProcSet[/PDF] Get a free answer to a quick problem. 0.297 Tc q 22.478 5.336 TD 1 Data in this Fast Fact represent the 50 states and the District of Columbia. /FormType 1 q ET >> Q /Length 54 98.843 5.203 TD 1.007 0 0 1.007 130.989 583.429 cm >> q /F3 12.131 Tf 421 0 obj Q >> 0.458 0 0 RG 0 G Q >> 1.005 0 0 1.007 102.382 726.464 cm endobj 89 0 obj /Length 63 1.014 0 0 1.006 111.416 437.384 cm /Resources<< /F3 17 0 R 1 i q endobj /BBox [0 0 88.214 16.44] /Resources<< 1 i 0 w Q (-20) Tj stream Q >> /F1 12.131 Tf endstream /Meta151 Do Q >> ET /BBox [0 0 88.214 35.886] /ItalicAngle 0 /Font << << Q Q 340 0 obj Q >> /FormType 1 /FormType 1 0 G 672.261 546.541 m 0 G /BBox [0 0 88.214 16.44] >> q 1.007 0 0 1.007 271.012 583.429 cm endobj /F4 12.131 Tf /Meta155 169 0 R ET q 1.007 0 0 1.007 45.168 713.666 cm stream endobj >> >> /FormType 1 /Subtype /Form /Type /Page >> 9.723 5.336 TD /ProcSet[/PDF] Q /Resources<< /BBox [0 0 88.214 16.44] /BBox [0 0 15.59 16.44] Q q q << /FormType 1 /ProcSet[/PDF] /Type /XObject /Meta41 55 0 R << >> /Type /XObject >> /Font << 1 i /F4 12.131 Tf stream Q Q >> /Type /XObject /Meta3 Do 1.007 0 0 1.007 654.946 347.046 cm (3) Tj Q /Resources<< endstream (4\)) Tj 0 g Q 549.694 0 0 16.469 0 -0.0283 cm /Meta281 Do 0 w endstream /ProcSet[/PDF] /Subtype /Form >> 0 G 0.564 G q /Subtype /Form q >> /Matrix [1 0 0 1 0 0] /BBox [0 0 549.552 16.44] 1 i /Meta320 334 0 R , Prove the following /Meta411 427 0 R stream q /FormType 1 q endstream /ProcSet[/PDF/Text] /XHeight 471 >> /Length 16 /Resources<< Q /Subtype /Form /Matrix [1 0 0 1 0 0] Q endstream /Length 245 1.005 0 0 1.007 102.382 546.541 cm /Meta51 Do 0 G 1.007 0 0 1.007 45.168 796.475 cm q stream 0 G 0 G Q stream 1.014 0 0 1.006 391.462 437.384 cm 1 i /F3 17 0 R q stream Q 0.458 0 0 RG << /BBox [0 0 15.59 16.44] 0 g /Matrix [1 0 0 1 0 0] /Type /XObject /Meta425 Do 0 g /Matrix [1 0 0 1 0 0] q Q Q /Length 59 /Matrix [1 0 0 1 0 0] >> 1.007 0 0 1.006 411.035 763.351 cm /Length 294 416 0 obj /FormType 1 Want to see the full answer? 1.007 0 0 1.007 411.035 849.172 cm endstream (D\)) Tj /Font << Q >> endobj Twice a number decreased by 8 gives 58. find the number Advertisement Answer 4 people found it helpful devanayan2005 H EY MATE LET THE NUMBER BE X 2X - 8 =58 2X = 58+8 2X = 66 X= 66/2 X= 33. >> Q /BBox [0 0 15.59 16.44] 0 20.154 m 51 0 obj /FormType 1 Q /Subtype /Form /Meta143 157 0 R Q /ProcSet[/PDF] /BBox [0 0 673.937 15.562] /Type /XObject /ProcSet[/PDF] This site is using cookies under cookie policy . q q /ProcSet[/PDF/Text] /F3 12.131 Tf stream /Meta60 Do 1 i endstream 1 i endstream /Font << /FormType 1 << endstream 352 0 obj q q >> /Length 118 1.502 24.649 TD Q /Length 16 /BBox [0 0 534.67 16.44] /Meta179 Do Q 0.486 Tc /Type /XObject /FormType 1 >> /FormType 1 0 w endobj endstream /Resources<< BT /Resources<< Q 0 g q 1.007 0 0 1.007 130.989 523.204 cm /Type /XObject /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /BBox [0 0 534.67 16.44] stream /FormType 1 >> Q /Length 245 /FormType 1 /Meta154 168 0 R /Meta70 Do /F3 12.131 Tf Q 0.737 w /Subtype /Form q >> Q 391 0 obj /Type /XObject 0 g Then the following equation can represent this problem: 17 + x = 68 We can subtract 17 from both sides of the equation to find the value of x. /Type /XObject 397 0 obj q /Type /XObject Q << << /F3 12.131 Tf Q endobj Q >> /ProcSet[/PDF/Text] /FontDescriptor 6 0 R Q 0.369 Tc 1 g q stream q q >> 0 G q endstream /F3 12.131 Tf << q 0 w q 1 i >> >> 0 w /Length 69 Q 0 g 16.469 5.203 TD >> /Length 69 Q endobj << (\)) Tj 0.458 0 0 RG q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1 g Q q 0.37 Tc /BBox [0 0 15.59 16.44] (ix) 3 less than 4 times a number is 17. Q << ET /BBox [0 0 15.59 16.44] /Type /Font q << Q /Type /XObject /Type /XObject Q /Meta405 421 0 R /ProcSet[/PDF] Q 1 i /Meta367 Do /FormType 1 1 i Q Q q /Resources<< /Font << << /Matrix [1 0 0 1 0 0] 0 4.894 TD q 0 G /Type /XObject << /FormType 1 Q 549.694 0 0 16.469 0 -0.0283 cm 1 i Phrase : Expression : 4 times some number : 4x: twice a number : 2y : one-third of some number : the product of a number and 12 : 12w: Some examples of common phrases and corresponding . stream endobj >> /F3 12.131 Tf /FormType 1 >> q /ProcSet[/PDF/Text] 0.458 0 0 RG , Prove the following /Resources<< >> q >> endstream 0 w Q /Matrix [1 0 0 1 0 0] /Font << endobj 20.21 5.203 TD /BBox [0 0 88.214 16.44] >> Q endstream >> (\)) Tj /FormType 1 q 1 g /Subtype /Form /Font << Q 44 0 obj >> /Type /XObject /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 85 0 obj /Font << /Meta147 161 0 R Q BT Q /Font << /Font << 219 0 obj /Resources<< >> /Meta393 Do stream stream /F3 12.131 Tf endobj q endstream 1 i q 1 g 385 0 obj << >> /Matrix [1 0 0 1 0 0] q q /Font << /BBox [0 0 15.59 29.168] q 1 i (1\)) Tj 0 w ET /Meta355 Do /Type /XObject endobj 1 g stream /ProcSet[/PDF/Text] >> stream /FormType 1 q q q /Meta10 Do 1.005 0 0 1.007 102.382 473.519 cm 1 g q 342 0 obj /Subtype /Form 430 0 obj 0 g (9) Tj /Resources<< 1 i Q endstream /Meta403 419 0 R q /Length 69 0 G q BT /Matrix [1 0 0 1 0 0] 0 5.203 TD 1 i /FormType 1 Q /F4 36 0 R 1 i /FormType 1 q endstream /Resources<< Q /Type /XObject /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] 0 G BT Q 0 g /ProcSet[/PDF/Text] >> /Meta368 382 0 R /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] endobj /Subtype /TrueType /Type /XObject /Ascent 1050 0 g /BBox [0 0 88.214 35.886] 0 g >> /Resources<< q /Font << /FormType 1 /Meta357 Do Q q BT /FormType 1 endobj /Subtype /Form /ProcSet[/PDF] /Resources<< endobj endobj -0.486 Tw /Meta423 439 0 R /F3 17 0 R q 424 0 obj (B) Tj /FormType 1 << ET /BBox [0 0 88.214 35.886] >> /F3 12.131 Tf /Meta291 305 0 R ET /FontDescriptor 16 0 R 0.564 G endobj /Meta7 Do Q BT q << 1 i (9\)) Tj 500 500 500 0 333 389 278 0 0 722 500 500]>> << 0 G 1.007 0 0 1.007 130.989 277.035 cm q Q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 130.989 277.035 cm /ProcSet[/PDF/Text] Q q 722.699 473.519 l q /Matrix [1 0 0 1 0 0] /Meta348 Do q endobj 2.238 5.203 TD q /FormType 1 << Q >> /ProcSet[/PDF/Text] /Meta371 385 0 R to represent the numbers. /Font << /ProcSet[/PDF] /Length 68 /ProcSet[/PDF/Text] 1 i 0 g q /FormType 1 /F3 12.131 Tf 0.838 Tc /FormType 1 0 G Q /ProcSet[/PDF/Text] /F3 17 0 R stream stream /Meta21 32 0 R endstream S /Resources<< BT endstream << Q Q /F4 12.131 Tf stream /AvgWidth 459 /F3 12.131 Tf 1.014 0 0 1.007 251.439 703.126 cm Kobe scored 85 points in a basketball game. (A\)) Tj /Meta140 Do endobj /Meta301 315 0 R ET 1.007 0 0 1.007 271.012 277.035 cm 0.524 Tc 0 G /Meta428 444 0 R Q stream /Type /XObject q /Resources<< /BBox [0 0 88.214 16.44] endstream q q 0.463 Tc 20.21 5.203 TD /BBox [0 0 88.214 35.886] /F1 7 0 R 0.241 Tc /Length 69 /Type /XObject stream /ProcSet[/PDF] /F3 12.131 Tf 6.746 5.203 TD 0 g q /Meta380 Do q >> /Resources<< 0 G q 0.458 0 0 RG /FormType 1 stream Q 0.737 w /Type /XObject 0 w << /Meta191 Do /Resources<< /Meta415 431 0 R /F3 17 0 R endobj q 0 G ET /Resources<< Q /BBox [0 0 88.214 16.44] ET 1 i Q endstream /ProcSet[/PDF] /ProcSet[/PDF/Text] >> 1 i 0.737 w 288 0 obj stream 354 0 obj 0.738 Tc BT endobj /Type /XObject q /Matrix [1 0 0 1 0 0] >> << Q q Q 0.738 Tc 0.564 G /Type /XObject q 0.458 0 0 RG /BBox [0 0 88.214 16.44] 679.036 293.596 m >> 1 i Q stream q endstream /ProcSet[/PDF] endstream /Type /XObject /Subtype /Form /Meta325 Do /Length 69 /Meta89 Do /ProcSet[/PDF] q 1.014 0 0 1.006 531.485 510.406 cm /Matrix [1 0 0 1 0 0] >> /FormType 1 1 g q >> Q BT q endobj stream (13) Tj 0.738 Tc 0.458 0 0 RG 1 i -0.486 Tw 0 5.203 TD 1.007 0 0 1.007 130.989 636.879 cm 58 decreased by twice Gails age. (D\)) Tj stream 1.014 0 0 1.006 391.462 836.374 cm 16.469 5.203 TD 0.737 w Q BT /Resources<< /Type /XObject q /ProcSet[/PDF/Text] ET /Font << 32.201 5.203 TD endstream /Font << Q 1 i 0 g stream q 3.742 5.203 TD endobj 0 w /F3 12.131 Tf /F3 17 0 R /Resources<< /Meta92 Do endobj Q /Type /XObject /Type /XObject /Meta98 Do q /F3 17 0 R q endobj 0 g /Length 54 79 0 obj >> >> /Subtype /Form Q >> /Subtype /Form /Meta326 340 0 R /Type /XObject /Matrix [1 0 0 1 0 0] /Resources<< q /Meta198 Do 1 i Q >> /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 347.046 cm /Meta127 141 0 R /Matrix [1 0 0 1 0 0] 0 5.336 TD >> 98 0 obj ET [(F)-22(ive)] TJ 1 i << /Resources<< 141 0 obj q >> /Matrix [1 0 0 1 0 0] q /Type /Pages endobj >> >> endstream q 1.007 0 0 1.007 45.168 746.789 cm VIDEO ANSWER: in this problem were asked to solve giving, given the following information. q >> 1 i 0.737 w 1.014 0 0 1.007 531.485 330.484 cm << /Subtype /Form 0 G 1.007 0 0 1.006 130.989 690.329 cm Q 0.564 G /Length 13 /FormType 1 0.564 G q endstream q /ProcSet[/PDF] 0 G Q q Q >> /Meta267 Do q /Type /XObject 407 0 obj Q q Q 1.502 5.203 TD endstream /Matrix [1 0 0 1 0 0] /Font << >> 1 i /Font << /Length 59 /F3 12.131 Tf Q q >> 0 G 0 w 0.458 0 0 RG >> >> 0 g /FormType 1 /ProcSet[/PDF/Text] q stream >> q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 0.68 Tc 0.564 G /Type /XObject 1 i endstream /Matrix [1 0 0 1 0 0] >> Q 20.21 5.203 TD /Resources<< ET /Length 118 /Meta277 Do endobj /Meta208 222 0 R 0 G /Flags 32 Q /F3 17 0 R /F3 12.131 Tf >> endobj /FormType 1 722.699 726.464 l /Type /XObject (D\)) Tj q 0 g -0.486 Tw /Resources<< /Resources<< /Meta290 304 0 R >> q >> /Meta146 160 0 R endstream 197 0 obj 1 i Q /BBox [0 0 534.67 16.44] /F3 12.131 Tf 0 G /Matrix [1 0 0 1 0 0] BT 295 0 obj >> stream 1 i endobj /F3 12.131 Tf Q What word phrase can you use to represent 5x + 2? >> ET q endobj >> /Length 16 Q >> q << /FormType 1 q 0.458 0 0 RG 0 g 1.007 0 0 1.007 271.012 583.429 cm endobj q /Subtype /Form Q 0 g >> /Meta193 207 0 R 224 0 obj 382 0 obj 0 g /Subtype /Form /Meta351 365 0 R /Type /XObject Q 1 i Q /Matrix [1 0 0 1 0 0] /FormType 1 /ProcSet[/PDF] 1.007 0 0 1.007 130.989 776.149 cm Q /ProcSet[/PDF/Text] 0 39.216 TD >> /Type /XObject Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. /Font << /Meta243 Do q /BBox [0 0 88.214 16.44] /Type /XObject Q 0.737 w q /Meta213 227 0 R /Length 54 /F3 17 0 R q 1 g Q endobj Q 1.007 0 0 1.007 411.035 330.484 cm q >> q Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 >> /Font << << q stream Q >> 1.007 0 0 1.007 130.989 330.484 cm /ProcSet[/PDF/Text] 1 i >> Two fewer than a number doubled is the same as the number decreased by 38. /Subtype /Form /Matrix [1 0 0 1 0 0] >> /FontDescriptor 35 0 R 0 g /Matrix [1 0 0 1 0 0] Q /ProcSet[/PDF] >> q Q /Resources<< /Length 54 >> /ProcSet[/PDF/Text] q ET (+) Tj /BBox [0 0 88.214 16.44] ET /Matrix [1 0 0 1 0 0] endobj Q /Matrix [1 0 0 1 0 0] 215 0 obj /Subtype /Form endobj endobj /Meta27 Do >> /Length 69 0.737 w endstream decreased by A number decreased by twelve X - 12 subtracted from Six subtracted from a number X - 6 Multiplication ( x ) times Eight times a number 8x the product of The product of fourteen and a number 14x twice; double Twice a number; double a number 2x multiplied by A number multiplied by negative six 6x 30.699 4.894 TD /Length 78 << /Length 2252 << /Type /XObject q stream endstream /Subtype /Form q << (9\)) Tj q q /Meta220 Do (+) Tj >> /Type /XObject /Font << 27.693 5.203 TD /Matrix [1 0 0 1 0 0] 0 w 0.564 G (2\)) Tj Q << /BBox [0 0 534.67 16.44] Q BT /Font << stream Q /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] /Meta13 Do stream 0 g /Length 118 endstream /F3 17 0 R 10.487 5.203 TD /Type /XObject endobj endobj Q /Meta20 31 0 R /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 130.989 636.879 cm /FormType 1 endobj << /Length 78 /Meta126 Do /Matrix [1 0 0 1 0 0] /Meta62 Do /Subtype /Form stream 0 4.894 TD Q Twice 4 bananas is 8. q BT endstream >> Q /F1 7 0 R q BT /Type /XObject Q 0 g >> /Type /XObject /Matrix [1 0 0 1 0 0] q /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Subtype /Form 0 w q Q /Type /XObject 43 0 obj /Matrix [1 0 0 1 0 0] /F3 17 0 R Q Q /Matrix [1 0 0 1 0 0] << q /Resources<< /BBox [0 0 88.214 16.44] q /Matrix [1 0 0 1 0 0] 311 0 obj << /Resources<< /Meta226 Do /Matrix [1 0 0 1 0 0] /Meta75 89 0 R q >> /Length 12 >> /FormType 1 /Meta195 209 0 R Q /Length 68 >> 332 0 obj 9 0 obj /Resources<< Q Q >> q (5\)) Tj Q )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] >> 0 w 1.005 0 0 1.007 79.798 713.666 cm /BBox [0 0 88.214 35.886] q 1 i /Length 16 /Meta426 442 0 R stream 1 i 23.952 4.894 TD