twice a number decreased by 58
twice a number decreased by 58
Mohegan Park Norwich Ct Fishing
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BT /Subtype /Form /FormType 1 /ProcSet[/PDF/Text] 0 g q 1 i q q /XHeight 447 >> /ProcSet[/PDF/Text] Q q Q BT 0 g /Meta22 33 0 R /FormType 1 /Font << q 0 G Q /Matrix [1 0 0 1 0 0] << ET /Subtype /Form /Matrix [1 0 0 1 0 0] /F3 17 0 R Q q endstream q q /Font << endstream /Subtype /Form Find an answer to your question Twice a number decreased by 8gives 58. /Meta66 Do /FormType 1 /Meta356 370 0 R stream q /F1 12.131 Tf /Meta384 Do << Q (11) Tj /Resources<< (-) Tj /Subtype /Form 1.007 0 0 1.007 130.989 277.035 cm ET /BBox [0 0 88.214 16.44] 125.064 4.894 TD endstream /Type /XObject 0.369 Tc 56 0 obj Q /Matrix [1 0 0 1 0 0] BT 549.694 0 0 16.469 0 -0.0283 cm /Meta311 325 0 R q /Matrix [1 0 0 1 0 0] stream /FormType 1 endobj BT 0 g endstream q >> Q q >> /Subtype /Form BT 2. BT BT BT 439 0 obj /Font << /Meta231 245 0 R 162 0 obj /Subtype /Form endobj >> endstream /Meta54 Do Q /Length 16 Q 0.17 Tc << 0 w /Meta214 228 0 R /F2 11 0 R /Resources<< /Subtype /Form /Meta97 111 0 R >> /Length 16 /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Resources<< Q 0.045 Tw /Resources<< /Meta53 67 0 R /Meta356 Do q q /ProcSet[/PDF/Text] /F3 17 0 R 0 g q >> 1 i /BBox [0 0 88.214 16.44] stream /Length 69 /CapHeight 476 q /Resources<< Q /F4 12.131 Tf 1.005 0 0 1.007 102.382 726.464 cm 38 0 obj ( x) Tj /Matrix [1 0 0 1 0 0] 94.364 5.203 TD /Resources<< (2) Tj Q 20.21 5.203 TD /Meta220 234 0 R ET /ProcSet[/PDF/Text] /Font << BT q 1.007 0 0 1.007 271.012 849.172 cm /Type /XObject /Type /XObject /F3 17 0 R << >> /FormType 1 1 i /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q 0 g 1 i -0.486 Tw 0 5.203 TD BT 1.007 0 0 1.006 551.058 836.374 cm 0 g /ProcSet[/PDF/Text] >> /Meta93 107 0 R >> endobj /Meta196 210 0 R q >> (D\)) Tj /ProcSet[/PDF/Text] endobj 0.564 G /Matrix [1 0 0 1 0 0] If LtitnS6S . /Resources<< /FormType 1 Q >> Q /Meta283 297 0 R /F3 17 0 R /Font << q q /Type /XObject endobj /FormType 1 /Subtype /Form Q Q /BBox [0 0 88.214 16.44] 1.502 24.339 TD BT Q endobj 0.68 Tc 0 g q /BBox [0 0 15.59 16.44] /Length 139 Q /Length 118 stream Q /ProcSet[/PDF/Text] Q /Matrix [1 0 0 1 0 0] /Resources<< 0 w q /FormType 1 /Matrix [1 0 0 1 0 0] Q 0 G /BBox [0 0 17.177 16.44] >> /Meta247 Do /ProcSet[/PDF/Text] endobj /FormType 1 Q q /Matrix [1 0 0 1 0 0] /Meta394 Do BT ET /BBox [0 0 15.59 16.44] >> /F4 12.131 Tf 231 0 obj q BT q 0 g /Length 54 1 i Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 654.946 400.496 cm >> /ProcSet[/PDF] /ProcSet[/PDF] Q BT endstream /Matrix [1 0 0 1 0 0] (\)) Tj BT /Length 16 /ProcSet[/PDF/Text] >> Q /Type /XObject Q /F3 17 0 R /F4 36 0 R 0.737 w >> stream /FormType 1 Q >> 1 i /Matrix [1 0 0 1 0 0] /Resources<< 3.742 5.203 TD /F3 12.131 Tf Q Q /Subtype /Form q 1 g Q /Resources<< 191 0 obj Q 0 G /Matrix [1 0 0 1 0 0] Example 1: Use the tables above to translate the following English phrases into algebraic expressions. /Font << /F3 17 0 R 0.564 G /Meta182 Do 0 w 1 g q 3.742 5.203 TD 0.737 w /ProcSet[/PDF/Text] << /F3 17 0 R >> /F3 17 0 R 19.474 5.203 TD stream Q 0 w >> << ET /BBox [0 0 88.214 16.44] q /Type /XObject >> /Subtype /Form Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 endobj 0.458 0 0 RG >> q 391 0 obj stream /F3 17 0 R /FormType 1 /Subtype /Form /FormType 1 q Q 1 i 1.007 0 0 1.006 130.989 690.329 cm /BBox [0 0 30.642 16.44] q 1 g /Meta285 Do /Meta34 Do 293 0 obj q [3] One half of a number increased by fourteen is twenty-one. 1.005 0 0 1.007 102.382 653.441 cm 0 w >> >> Q q /Font << >> /ProcSet[/PDF/Text] 1 i q Q stream 0 G /Resources<< /Font << endobj /Resources<< q endstream /Meta394 410 0 R ET Q Question. << q /Meta156 170 0 R BT q Q /F1 7 0 R q Q /Meta275 Do >> Q 1.005 0 0 1.009 45.168 905.633 cm /Resources<< /Resources<< 0 g BT 0 w q /Type /Page 1.014 0 0 1.007 111.416 776.149 cm /F4 12.131 Tf 0 g q /Meta123 137 0 R q /ProcSet[/PDF/Text] Q 0 g 0 g /ProcSet[/PDF/Text] /Resources<< Q /BBox [0 0 15.59 29.168] (D\)) Tj stream /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 813.037 cm endstream /Resources<< q stream endstream >> 0 w 0.737 w /FormType 1 Q /FormType 1 /ProcSet[/PDF/Text] /Meta142 156 0 R 1.007 0 0 1.007 271.012 636.879 cm /Subtype /Form >> Q [(Fiv)25(e ti)18(me)16(s)] TJ q ET /BBox [0 0 15.59 29.168] /Length 59 Q /Meta112 Do /Font << /Meta334 348 0 R 1.007 0 0 1.007 130.989 523.204 cm 0 g >> /ProcSet[/PDF/Text] q /F3 12.131 Tf 285 0 obj stream >> /Meta422 438 0 R 0 g /Matrix [1 0 0 1 0 0] 128 0 obj << << /Subtype /Form endstream << q /Matrix [1 0 0 1 0 0] endobj 296 0 obj /Font << /Length 69 0 g endobj /Meta144 158 0 R endstream 0 5.203 TD endobj q << 1.007 0 0 1.007 271.012 776.149 cm /BBox [0 0 15.59 16.44] /Font << stream >> endobj q q /ProcSet[/PDF/Text] /Meta13 Do /BBox [0 0 88.214 16.44] q 252 0 obj /ProcSet[/PDF] /Length 59 Q 0 w >> /Type /XObject q /Length 16 /MissingWidth 250 1 i 1 g BT endobj 1 g /F3 17 0 R 1.014 0 0 1.007 391.462 523.204 cm 0.564 G /XObject << Q /Resources<< /Length 54 /FormType 1 << q Q 309 0 obj Q Q q >> /BBox [0 0 88.214 35.886] Q -0.062 Tw /F3 12.131 Tf /FormType 1 /Matrix [1 0 0 1 0 0] 287 0 obj /Type /XObject 1 g >> >> 19 0 obj /Font << /Resources<< >> 1 i /F3 12.131 Tf q >> /F4 12.131 Tf /Meta180 194 0 R 0 w Q ET stream endobj /FormType 1 /BBox [0 0 534.67 16.44] 99 0 obj endobj /Resources<< /Meta8 Do q /XObject << /Length 69 Q /Type /XObject /Type /XObject >> /Subtype /Form 0 G /FormType 1 >> ET 282 0 obj /Font << /Matrix [1 0 0 1 0 0] >> /BBox [0 0 88.214 16.44] /Subtype /Form 1.014 0 0 1.007 251.439 703.126 cm Q /Meta51 Do /Length 78 /Length 69 1 i 1 i 0 G stream Q 236 0 obj Q q /BBox [0 0 88.214 35.886] Q /Subtype /Form 0 G endobj /ProcSet[/PDF/Text] 0 g endobj 50 0 obj Q q Q 0.425 Tc 1.014 0 0 1.007 251.439 583.429 cm /Meta3 12 0 R /Type /XObject /Matrix [1 0 0 1 0 0] Q Q 259 0 obj 1 i 0.458 0 0 RG 159 0 obj /F3 12.131 Tf 415 0 obj >> q /Meta183 197 0 R endobj /Matrix [1 0 0 1 0 0] /Type /XObject q /ProcSet[/PDF] /ProcSet[/PDF] ET Q (x) Tj endobj -0.463 Tw ET 0 g 0.564 G /Length 64 q 1 i 0 G Q q Q /BBox [0 0 30.642 16.44] /Length 59 [( the )-24(sum of a n)-14(umber an)-14(d )] TJ ET 90 0 obj /Type /XObject 1 i Q endstream q 23.952 4.894 TD stream 1 i /Subtype /Form /Meta399 415 0 R q Q >> /F3 12.131 Tf Q /Meta229 Do /Meta373 Do 16.469 5.203 TD q /ProcSet[/PDF] /Meta382 Do q q /FormType 1 /Subtype /Form /Matrix [1 0 0 1 0 0] /FormType 1 9.723 5.336 TD /Meta296 Do stream 0.564 G Q q ET Q endobj 0 g Q /Matrix [1 0 0 1 0 0] /Type /XObject /Subtype /Form Q << 67 0 obj stream 1.007 0 0 1.007 411.035 636.879 cm /Resources<< /Meta98 Do 0.564 G /Font << /Subtype /Form /BBox [0 0 15.59 16.44] >> 0 g endobj 1 g endobj 0.271 Tc q /FontDescriptor 35 0 R 388 0 obj 1.007 0 0 1.007 130.989 776.149 cm q 1.005 0 0 1.007 102.382 799.486 cm /F3 12.131 Tf 0 0 Similar questions Find the number which when decreased by 8% becomes 506. /F1 12.131 Tf Step 1/1. 0.564 G /Type /XObject ET 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. [(-1)-16(52)] TJ q /Resources<< /ProcSet[/PDF] endstream Q 0 G gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. /BaseFont /PalatinoLinotype-Bold /Subtype /Form /FormType 1 /FormType 1 /Length 63 << /Count 2 Q /Type /XObject Q /Length 69 << /Meta134 148 0 R q 367 0 obj 0.458 0 0 RG 0.524 Tc q /Matrix [1 0 0 1 0 0] << /Type /XObject ET /Resources<< >> >> endobj /Meta172 Do /Resources<< stream /Type /XObject /F3 17 0 R /Resources<< /Length 64 0 g 0 5.203 TD >> 0 5.203 TD /Matrix [1 0 0 1 0 0] Q Q /FormType 1 q Q /Resources<< /Length 69 q q 16.469 5.336 TD q << /Length 65 q /FormType 1 endstream ET Q /Resources<< >> /Meta270 Do >> 0.458 0 0 RG Q 0.786 Tc 24 0 obj /FormType 1 0.458 0 0 RG /FormType 1 0 g endstream /Length 69 /FormType 1 /FormType 1 >> S /Length 68 /Matrix [1 0 0 1 0 0] /Meta32 Do 0 g 9.723 5.336 TD 234 0 obj 0.458 0 0 RG Q /Resources<< /ProcSet[/PDF/Text] /Resources<< 0 5.203 TD /Meta365 379 0 R /F3 17 0 R BT endobj Q >> /Type /XObject /Matrix [1 0 0 1 0 0] 69 0 obj stream /F3 12.131 Tf Q Q q 2.238 5.203 TD stream /Subtype /Form endobj Twice a number decreased by ten is greater than 24. 402 0 obj 408 0 obj endobj Q endstream >> /Length 69 0.458 0 0 RG /Resources<< /ProcSet[/PDF/Text] 118 0 obj /Subtype /TrueType ET endobj /F3 17 0 R 173 0 obj Q /Type /XObject The sum Of twice a nu4ber What is the number? q (C) Tj BT 89 0 obj /BBox [0 0 639.552 16.44] >> /Type /XObject /Type /XObject endobj /FormType 1 0 g Q /F3 12.131 Tf /Font << 0 g q /Matrix [1 0 0 1 0 0] q 0 g 0 g 1 i /Subtype /Form -0.106 Tw /BBox [0 0 88.214 16.44] /Meta5 14 0 R 0 g /FormType 1 /Length 78 >> >> 0 g >> 0 20.154 m ET stream /Meta219 233 0 R q /BBox [0 0 88.214 16.44] 32.201 5.203 TD /Resources<< 1.005 0 0 1.007 102.382 546.541 cm >> Q 225 0 obj << q (x ) Tj /F3 17 0 R Q /Type /XObject Q /Meta203 Do ET )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] << /Matrix [1 0 0 1 0 0] >> 1 0 obj /FormType 1 0 G q /Type /XObject << >> 0 g If a number is 50%, then it is a half - the same as 0.5 or 1/2. /Font << 130 0 obj /Meta197 211 0 R q BT endobj 1 i q /ProcSet[/PDF/Text] q Q ET /Meta180 Do /Matrix [1 0 0 1 0 0] q 0.838 Tc /Meta48 Do /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] 1 i >> endobj /Subtype /Form /Matrix [1 0 0 1 0 0] stream 1 i >> /Meta23 Do /Length 69 /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /Subtype /Form 1.007 0 0 1.007 271.012 330.484 cm /ProcSet[/PDF/Text] 1 i /Meta388 404 0 R q 0 w /ProcSet[/PDF/Text] Q Q Q endobj >> 0 G 1 i /Resources<< >> 202 0 obj Q /BBox [0 0 88.214 35.886] 0 g >> Q >> >> 0 g /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] Q >> << /ProcSet[/PDF/Text] /Type /XObject >> /BBox [0 0 30.642 16.44] /Font << /Type /XObject /Subtype /Form /Resources<< 0.564 G /F3 17 0 R Q /FormType 1 0 g /Meta307 321 0 R /Matrix [1 0 0 1 0 0] stream q /F1 12.131 Tf /Size 447 /Subtype /Form /Resources<< q /Length 68 0.786 Tc Q q /Meta174 188 0 R q In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). Q The value of k is: (b) 3 (d) 0 (a) 4 (c) -4 TL ing:, 1)take a graph and draw two perpendicular lines to obtain four uadrants 2)draw any object using straight line 3) write the coordinates of each point o /Meta98 112 0 R 0 g q >> /F3 12.131 Tf Q /Meta111 Do 0 G q 0 g endstream 8 0 obj Q q /Length 118 endstream Notice that we used the variable \large {d} d in our equation to stand for our unknown value. 1.502 5.203 TD /BBox [0 0 549.552 16.44] /Type /XObject ( x) Tj /Meta169 183 0 R endstream /Font << /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q Q /ColorSpace [/Indexed /DeviceGray 1 ] /Type /XObject /Type /FontDescriptor q 0.738 Tc /Flags 32 /Resources<< /Meta327 Do 97 0 obj 1.007 0 0 1.007 411.035 277.035 cm /FormType 1 1 i stream 1 g ET endobj 0.838 Tc BT q q BT /Matrix [1 0 0 1 0 0] 0 G 0.311 Tc endstream Q /Length 59 /FormType 1 /Type /XObject 1 i 1 i /Subtype /Form 0.524 Tc Q /F3 12.131 Tf /ProcSet[/PDF] q 444 0 obj 290 0 obj 1.014 0 0 1.007 531.485 849.172 cm Q << /BBox [0 0 88.214 16.44] endobj BT >> /Meta254 Do 0.838 Tc 1 i q /ProcSet[/PDF] >> /Subtype /Form Q /Length 69 Q q /F1 7 0 R 0.838 Tc /Length 16 0 g 0000000000 65535 f 0000140665 00000 n 0000140732 00000 n 0000000015 00000 n 0000120613 00000 n 0000000126 00000 n 0000000314 00000 n 0000000577 00000 n 0000001009 00000 n 0000001360 00000 n 0000001548 00000 n 0000001817 00000 n 0000002237 00000 n 0000002627 00000 n 0000002815 00000 n 0000003222 00000 n 0000003409 00000 n 0000003679 00000 n 0000004128 00000 n 0000004396 00000 n 0000004636 00000 n 0000004952 00000 n 0000005229 00000 n 0000005533 00000 n 0000005720 00000 n 0000005984 00000 n 0000006171 00000 n 0000006438 00000 n 0000006678 00000 n 0000006931 00000 n 0000007188 00000 n 0000007375 00000 n 0000007642 00000 n 0000007882 00000 n 0000008135 00000 n 0000008318 00000 n 0000008578 00000 n 0000008755 00000 n 0000009012 00000 n 0000009280 00000 n 0000009467 00000 n 0000009734 00000 n 0000009974 00000 n 0000010247 00000 n 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00000 n 0000129726 00000 n 0000129914 00000 n 0000130183 00000 n 0000130372 00000 n 0000130627 00000 n 0000130815 00000 n 0000131084 00000 n 0000131273 00000 n 0000131528 00000 n 0000131716 00000 n 0000131985 00000 n 0000132174 00000 n 0000132429 00000 n 0000132617 00000 n 0000132886 00000 n 0000133075 00000 n 0000133330 00000 n 0000133518 00000 n 0000133787 00000 n 0000133976 00000 n 0000134231 00000 n 0000134419 00000 n 0000134688 00000 n 0000134877 00000 n 0000135132 00000 n 0000135320 00000 n 0000135589 00000 n 0000135778 00000 n 0000136033 00000 n 0000136221 00000 n 0000136499 00000 n 0000136688 00000 n 0000136943 00000 n 0000137186 00000 n 0000137447 00000 n trailer /Meta352 Do Q /Length 69 /Subtype /Form endstream >> BT /BBox [0 0 88.214 16.44] /Subtype /Form endstream /FormType 1 0 g 1.005 0 0 1.007 102.382 599.991 cm 0 g 0 g /ProcSet[/PDF/Text] A rectangular garden has a width that is 8 feet less than twice the length. << q << -0.486 Tw /F3 17 0 R 20.21 5.203 TD Answer link. Q /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] /ItalicAngle 0 /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] >> 1 i Q BT 1 i q (D) Tj /Meta29 42 0 R q 220 0 obj Q q /Type /XObject 1.007 0 0 1.006 411.035 437.384 cm /Type /XObject /Subtype /Form q /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 1.007 0 0 1.007 271.012 383.934 cm /Type /XObject >> << /F1 12.131 Tf 24.718 8.18 TD 1.502 7.841 TD 0 5.203 TD /Length 69 endobj Q /Length 16 Q BT /ProcSet[/PDF/Text] /Length 68 q 380 0 obj 22.478 5.203 TD ET 1.014 0 0 1.006 251.439 437.384 cm /Length 54 >> /BBox [0 0 30.642 16.44] 0 G /FormType 1 stream stream BT /Subtype /Form 0.458 0 0 RG /Resources<< /ProcSet[/PDF/Text] (6\)) Tj q q endobj << stream 1.014 0 0 1.007 391.462 450.181 cm >> 0.369 Tc 1.007 0 0 1.007 130.989 636.879 cm /ProcSet[/PDF] /Subtype /Form /Meta328 Do >> endobj 0.458 0 0 RG q /Type /XObject >> >> /F3 12.131 Tf 274 0 obj /BBox [0 0 17.177 16.44] 0 g >> BT /F3 17 0 R /Length 16 /I0 Do q /Subtype /Form 1.007 0 0 1.007 551.058 383.934 cm /Subtype /Form 1.007 0 0 1.007 271.012 450.181 cm /Meta84 Do /BBox [0 0 88.214 16.44] /Subtype /Form endstream /Resources<< /Meta258 Do A. /Font << /Matrix [1 0 0 1 0 0] 1 i /Font << /F3 17 0 R Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM Q endstream BT 0 G /ProcSet[/PDF/Text] stream /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 271.012 776.149 cm /FormType 1 Q /Length 118 >> /Matrix [1 0 0 1 0 0] Q 17 0 obj Q Q BT Q /Font << >> 0 G q ET /Meta389 Do 1 i >> Q stream q >> /Meta316 Do 1 i /Subtype /Form Q stream Q /Meta21 32 0 R q Q (x ) Tj /Type /XObject /ProcSet[/PDF/Text] 1 i /Meta357 Do /Font << /Subtype /Form /Meta149 163 0 R /Font << q Q /Subtype /Form << >> endstream >> BT q /Meta361 Do /Type /XObject /ProcSet[/PDF/Text] 0.737 w /Length 58 /Type /XObject q 43 0 obj /Resources<< 1 g /Subtype /Form Q /Meta267 Do /FormType 1 1.014 0 0 1.006 391.462 836.374 cm /ProcSet[/PDF/Text] Q << 1 i 0 4.894 TD q /Descent -299 /Matrix [1 0 0 1 0 0] Q 1 g q 0 g 0.564 G ET q stream Q /Type /XObject 0 20.154 m 0.737 w /Meta241 Do /Meta237 251 0 R BT Q /F4 12.131 Tf /Font << 0 G q [(The )-19(quotient of )] TJ << Q /FormType 1 q stream /ProcSet[/PDF] /Meta125 139 0 R >> endobj /Resources<< Q >> << ET /Meta294 Do >> q Q BT stream /Font << Q BT Q << q /Subtype /Form /Font << Q /ProcSet[/PDF] /F3 12.131 Tf (2) Tj >> Q /F3 12.131 Tf BT 1 g Q BT /Resources<< /F3 12.131 Tf endstream [tex]\sin (\pi -x)=\sin x[/tex]. endobj 384 0 obj /Length 69 0 g /Meta360 Do q q /Meta23 34 0 R /Matrix [1 0 0 1 0 0] Q q BT 1 i /Resources<< /F4 36 0 R /Type /XObject 1 g /FontBBox [-170 -292 1419 1050] ET q /Type /XObject q >> Q /F3 12.131 Tf /F3 12.131 Tf /Subtype /Form /BBox [0 0 88.214 16.44] q Q Q Q /Subtype /Form >> /BBox [0 0 30.642 16.44] /Resources<< Q Q 1 i /F3 12.131 Tf 0.737 w << 1.007 0 0 1.007 411.035 636.879 cm >> 1 i >> >> Q 0.486 Tc /Meta70 Do [(th)-28(e di)-18(ffe)-14(ren)-23(ce o)-28(f )] TJ q q xref q 0.564 G /FormType 1 /Length 59 0.51 Tc 0.737 w 326 0 obj << << /Matrix [1 0 0 1 0 0] endstream /Matrix [1 0 0 1 0 0] 197 0 obj q 1 i /BBox [0 0 88.214 35.886] /Resources<< /FormType 1 (\(x ) Tj /F3 12.131 Tf 1.007 0 0 1.007 130.989 523.204 cm /F3 12.131 Tf endobj /Meta54 68 0 R 1.007 0 0 1.007 551.058 636.879 cm q 0 g q /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Subtype /Form 1 g q q 39 0 obj 0 g Q 1 i . Q stream Q q /Font << ET /Type /XObject /FormType 1 (+) Tj q 14.966 20.154 l /FirstChar 32 /Type /XObject 0.68 Tc The results found were expressed mainly through tables and graphs as the main resources of the statistical language. /F3 12.131 Tf Q Q endstream q Q 1.007 0 0 1.007 45.168 730.228 cm (-) Tj 20/n b.) /BBox [0 0 673.937 16.44] 0 w 1.007 0 0 1.007 551.058 330.484 cm Q /Type /XObject /Length 16 1 i /F1 7 0 R ET /Meta304 318 0 R >> /Type /XObject q 5 0 obj Q endstream q 0 g >> >> /Subtype /Form endstream Q 1.007 0 0 1.006 411.035 690.329 cm q 1.007 0 0 1.007 45.168 763.351 cm BT q -0.029 Tw Q >> Q >> /Meta376 390 0 R BT stream /F3 17 0 R /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] << 0 g /ProcSet[/PDF] /BBox [0 0 88.214 16.44] endobj stream [(The )-19(quotient of )] TJ /Meta358 372 0 R q /Subtype /Form 1 i /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 411.035 330.484 cm 0 g endobj << -0.486 Tw /Font << /BBox [0 0 15.59 16.44] /Meta9 Do /Subtype /Form /Matrix [1 0 0 1 0 0] &K @ /Meta277 Do 1 i 1 i 1.007 0 0 1.007 551.058 277.035 cm 0 833 610 0 0 0 667 778 0 1000 0 0 0 0 0 0 << 0.458 0 0 RG 1.007 0 0 1.007 45.168 763.351 cm << Q Q Q 0 w /Meta274 288 0 R << >> endstream ET endobj /Length 294 /Subtype /Form 1 i 0.458 0 0 RG /FormType 1 << >> D. Twice a number decreased by ten is less than 24. /Length 69 Q q endstream q (iii) 25 exceeds a number by 7. /F3 17 0 R /Resources<< 1.014 0 0 1.007 531.485 636.879 cm 0 G >> 332 0 obj BT Q /ProcSet[/PDF/Text] /Resources<< /Resources<< 30 0 obj (x) Tj endstream q Q C. Twice a number decreased by ten is at most 24. 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 /Type /XObject 0.369 Tc /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] >> q /Resources<< 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 . q q << /Meta214 Do 0 G /Meta63 Do /FormType 1 1 i (58) Tj /Subtype /Form q Q 0.737 w 0 w /Matrix [1 0 0 1 0 0] /Length 88 /FormType 1 >> Q /Font << endobj Q Q /Resources<< /F3 17 0 R (5\)) Tj Q /Font << 0.737 w endstream Q >> /FormType 1 0 5.203 TD Q Q /F3 17 0 R q Q Six subtracted from a number 6. 1 i stream Q /BBox [0 0 88.214 16.44] ET ET 1 i /BBox [0 0 15.59 16.44] 1 i q /Length 57 0 w ET Easy Solution Verified by Toppr Let the number be x. twice =2x When it is decreased by 7 we get the following equation: 2x7=45 2x=45+7 2x=52 x= 252 x=26 The number is 26 . /Length 16 endobj 1.007 0 0 1.007 654.946 347.046 cm /Meta404 Do stream /Resources<< /Type /XObject /Meta264 278 0 R /FormType 1 /BBox [0 0 88.214 16.44] 0 G 1.014 0 0 1.007 251.439 277.035 cm Q (x) Tj << >> << stream /Length 12 1 g /FormType 1 q q /Type /XObject 0 w /Length 59 ET 1 i /Subtype /Form /Length 294 (B\)) Tj >> Q q stream Q Q Q /Length 16 q endobj /Type /XObject q 1.014 0 0 1.006 251.439 836.374 cm 1.005 0 0 1.006 45.168 879.284 cm q >> /F3 17 0 R /Meta92 106 0 R /Font << 0 g /ProcSet[/PDF/Text] stream /F3 17 0 R /Subtype /Form >> q Q /Font << 1 i >> /Matrix [1 0 0 1 0 0] /Length 69 endobj Q q q /Subtype /Form 0.369 Tc 221 0 obj ET 0 w /Font << 0 G /F3 12.131 Tf Q Q 0 g /Subtype /Form /Length 69 /Resources<< [(Testnam)19(e: 1.12 T)16(RAN)16(SLATING )17(ALG EXPRES)21(SIONS 2)] TJ /F3 12.131 Tf 1.007 0 0 1.007 551.058 703.126 cm /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] /Length 70 >> 0 G Q /Meta355 Do 72 0 obj 1 i Q /FormType 1 2.238 5.203 TD stream /F1 12.131 Tf stream endobj /Subtype /Form 0 g Q Q /F1 12.131 Tf q Q endstream >> << 1.502 5.203 TD /Resources<< BT /Matrix [1 0 0 1 0 0] /Resources<< Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. /Length 69 >> 1.014 0 0 1.007 391.462 636.879 cm q /Meta314 Do q stream BT q << Q q q 1.008 0 0 1.007 654.946 293.596 cm /Length 16 0 G -0.463 Tw >> Q /ProcSet[/PDF/Text] endstream -y. Q /Meta294 308 0 R q 0.458 0 0 RG 0 G /Resources<< q Diabetes, also known as diabetes mellitus, is a group of common endocrine diseases characterized by sustained high blood sugar levels. Q q ET 366 0 obj q 20.21 5.203 TD >> /Meta362 376 0 R ET 1.007 0 0 1.007 67.753 473.519 cm 0 g Q >> /Font << /Meta269 Do /Type /Font /Meta206 220 0 R endstream /BBox [0 0 30.642 16.44] Q /Type /XObject /Matrix [1 0 0 1 0 0] BT 0.024 Tw q Q /Matrix [1 0 0 1 0 0] /Subtype /Form 288 0 obj /Length 16 15.731 5.336 TD /FormType 1 1.014 0 0 1.007 531.485 523.204 cm 0 4.894 TD 233 0 obj /Length 106 57.656 5.203 TD Q Q endstream Q /Meta274 Do 0.564 G 0 g Q /Font << Q BT /Resources<< 0.68 Tc /Matrix [1 0 0 1 0 0] /Subtype /Form q /Length 151 q 0.524 Tc /Type /XObject /Meta323 Do /Matrix [1 0 0 1 0 0] endobj 430 0 obj Q Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] /F1 12.131 Tf /ProcSet[/PDF] /F3 12.131 Tf q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 551.058 636.879 cm >> the quotient of twenty and a number a.) /Matrix [1 0 0 1 0 0] >> (C\)) Tj /ProcSet[/PDF/Text] /Type /XObject /Meta249 Do Q 1.007 0 0 1.007 130.989 383.934 cm BT >> BT 275 0 obj 2 0 obj q stream /Length 16 /Resources<< 0.564 G 19.474 5.203 TD /Matrix [1 0 0 1 0 0] endstream 1.007 0 0 1.007 551.058 703.126 cm /ProcSet[/PDF/Text] >> endobj Q 722.699 799.486 l q 272 0 obj 0.369 Tc twice - means that a number (the unknown value) is multiplied by 2 2 With these in mind, let's write our algebraic equation. << /Meta79 93 0 R 360 0 obj >> /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] endobj 1 g ET ET /Matrix [1 0 0 1 0 0] 1 i q /FormType 1 /Resources<< ET ET Q << BT q (B\)) Tj Q /Font << q /F4 36 0 R >> Q 0.564 G 1.007 0 0 1.007 67.753 599.991 cm
twice a number decreased by 58