1 i 0.68 Tc /Resources<< 0 G 1 i /Font << 0.458 0 0 RG /Matrix [1 0 0 1 0 0] /Meta325 Do /Meta53 67 0 R /Resources<< Q /Meta417 Do /ProcSet[/PDF/Text] /ItalicAngle 0 q 1.502 5.203 TD /FormType 1 Q /ProcSet[/PDF/Text] Q << << 337 0 obj 1.014 0 0 1.006 111.416 437.384 cm >> q 246 0 obj 0 g >> /Type /XObject << endobj 1.007 0 0 1.007 654.946 347.046 cm /Subtype /Form << 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. >> 1 i /Meta271 285 0 R /F4 36 0 R Q /Subtype /Form /Meta188 Do << endobj /F1 7 0 R >> 1.014 0 0 1.007 111.416 277.035 cm /ProcSet[/PDF/Text] 315 0 obj /BBox [0 0 88.214 16.44] /Meta280 294 0 R /BBox [0 0 673.937 15.562] 0.738 Tc >> q << /Matrix [1 0 0 1 0 0] endstream << /Type /XObject 0.227 Tc /Matrix [1 0 0 1 0 0] >> >> 20.21 5.203 TD >> /Type /XObject /Resources<< /Meta290 Do Q << << q 1 g 101 0 obj ( x) Tj q 3.742 5.203 TD BT /Resources<< Q Q Q 1 i /ProcSet[/PDF] (x) Tj q Q /Meta221 235 0 R /ProcSet[/PDF/Text] /Length 16 0 g ET >> q 9.723 5.336 TD Percent Change = (Decrease First Value) x 100% Q /F3 12.131 Tf /Type /XObject Q BT /F4 36 0 R 0 w 0 g 1.014 0 0 1.006 111.416 690.329 cm 1 i /Meta416 Do Q Q /Matrix [1 0 0 1 0 0] /Type /XObject >> 1 i /F3 17 0 R Q Q >> q /Length 69 /FormType 1 Q q /FormType 1 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. /Type /XObject /ProcSet[/PDF/Text] /ProcSet[/PDF] 1.007 0 0 1.007 654.946 546.541 cm 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 /F4 12.131 Tf q 11 0 obj ET Q /Subtype /Form >> q /Length 69 << 1 g ET >> >> Q /Meta227 241 0 R /Meta206 220 0 R q >> << >> /BBox [0 0 17.177 16.44] 0 G /Matrix [1 0 0 1 0 0] q Q /ProcSet[/PDF/Text] /Resources<< /Matrix [1 0 0 1 0 0] << /Matrix [1 0 0 1 0 0] endobj 305 0 obj >> /Resources<< << 1.007 0 0 1.007 271.012 523.204 cm /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /FormType 1 >> /Meta102 116 0 R /F3 17 0 R endobj 0 G 0.458 0 0 RG /ProcSet[/PDF/Text] /Meta54 68 0 R /F3 17 0 R q /BBox [0 0 534.67 16.44] q 363 0 obj /Meta98 112 0 R /ProcSet[/PDF] Q q 24 0 obj /AvgWidth 657 q /FormType 1 endstream endobj Q /Font << 1.014 0 0 1.007 391.462 583.429 cm q Patients' reasons for declining screening were not collected . /Length 68 >> /Length 16 saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. stream q Q stream 35,000 worksheets, games, and lesson plans, Spanish-English dictionary, translator, and learning, **Note: You could choose any variable you want, to represent the numbers. /Type /XObject 0 G /Resources<< /Font << 165 0 obj 0.564 G << /BBox [0 0 88.214 35.886] /Meta284 Do /Meta417 433 0 R q 35.206 4.894 TD /Length 16 startxref /Matrix [1 0 0 1 0 0] /Subtype /Form Q ET << /Resources<< >> /Matrix [1 0 0 1 0 0] 0 G /Matrix [1 0 0 1 0 0] 0.737 w >> 0 g >> >> q /F3 12.131 Tf /Length 16 Q 0.838 Tc 0.463 Tc /Font << stream /FormType 1 0 G /Meta235 249 0 R /Font << endobj << >> endstream /Subtype /Form q >> /Meta310 324 0 R << ET stream 1.007 0 0 1.006 130.989 690.329 cm ET q 3.742 8.18 TD /Resources<< q q /FormType 1 q 4 0 obj Q >> /Meta238 252 0 R /I0 51 0 R (-11) Tj >> endobj endstream 1 i q /Meta174 Do Q Q ET endobj stream /Subtype /Form /Length 60 /Resources<< << (x) Tj /BBox [0 0 88.214 16.44] /BBox [0 0 17.177 16.44] >> /F3 12.131 Tf q << ET q >> Q /Length 16 1.007 0 0 1.006 130.989 690.329 cm /FormType 1 >> (B) Tj 1 i >> ET /F3 12.131 Tf /Meta149 163 0 R /Type /XObject /Type /XObject /Resources<< Q /Font << 0 g >> /Meta262 276 0 R BT q /Meta197 Do 250 0 obj /Matrix [1 0 0 1 0 0] 0 G /ProcSet[/PDF/Text] q /Length 95 >> /FormType 1 /BBox [0 0 17.177 16.44] 36 0 obj >> /Meta187 Do (x ) Tj Choose the correct one. q /Length 16 Q /Font << /Length 2252 /BBox [0 0 88.214 16.44] Q endstream 1.007 0 0 1.007 551.058 703.126 cm /Type /XObject 0.737 w /Subtype /Form /Width 734 1 i endstream BT q q /Resources<< q /Meta293 307 0 R 0 g /Resources<< 21.713 20.154 l /I0 Do /Subtype /Form /CapHeight 476 0 G Q /FormType 1 >> 1 i ET 1 i /Resources<< q 388 0 obj /ProcSet[/PDF/Text] >> Q q Q 135 0 obj /Resources<< >> >> 0.737 w >> /BBox [0 0 88.214 16.44] q q /Font << 1.005 0 0 1.007 79.798 746.789 cm 0.564 G Q 1.005 0 0 1.007 102.382 653.441 cm /Subtype /Form /BBox [0 0 88.214 35.886] >> 0 g stream 0.458 0 0 RG /Length 81 /Length 67 q /Matrix [1 0 0 1 0 0] /Resources<< q /Subtype /Form Q /Resources<< 13.464 5.203 TD q >> Q Q /Resources<< 76.394 5.203 TD /Type /XObject /Matrix [1 0 0 1 0 0] /Type /XObject /Meta178 Do >> /ProcSet[/PDF/Text] 0 G Q q /Meta312 326 0 R >> >> q Q Q << Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. /F3 17 0 R /ProcSet[/PDF/Text] << 15 0 obj C. Twice a number decreased by ten is at most 24. 0 g >> /FormType 1 q >> Q /Resources<< >> q /ProcSet[/PDF/Text] 32.939 5.203 TD /BBox [0 0 17.177 16.44] Thirthy is equal to twice a number decreased by four = solve and check the equation? Q stream ET 0.737 w /Font << 0.564 G endobj endobj 177 0 obj (9) Tj /F4 12.131 Tf q endstream Q /Subtype /Form >> /F3 12.131 Tf q 0 G /Subtype /Form 0 g endstream stream 0.564 G stream /F3 12.131 Tf /Meta256 Do /BBox [0 0 88.214 16.44] ET /Subtype /Form [(1.1)21(2 Tran)36(sla)18(tin)23(g Alge)17(b)26(raic )18(Exp)22(res)24(si)25(on 2)] TJ /Length 88 0.458 0 0 RG Q 1.007 0 0 1.006 411.035 437.384 cm endobj /Meta217 Do /Subtype /Form endstream Q /FormType 1 endobj endstream BT stream q 1.005 0 0 1.007 102.382 490.08 cm >> 7) The quotient of 40 and the product of a number and -8 7) A) 40 x - 8 B) -320 x C) 40-8x D)-8x 40 8) Twice a number, decreased by 58 8) A) 2 (x - 58 ) B) 2 x - 58 C) 2 x + 58 D) 2 (x + 58 ) 9) A number subtracted from -20 9) A) -20 x B) -20 + x C) x - (-20 ) D) -20 - x 10) Five times the sum of a number and -23 10) /FormType 1 The solution of the equation ax + b = 0 is Solution: (c) The equation is ax + b = 0 ax - b Solution is Question 2. /Resources<< /Resources<< /Matrix [1 0 0 1 0 0] /FormType 1 /F3 12.131 Tf >> Kobe scored 85 points in a basketball game. Q << 1.007 0 0 1.007 654.946 599.991 cm a Question q q stream 0.737 w 0 g 113 0 obj ET << /BBox [0 0 534.67 16.44] >> 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. /Subtype /Form /BBox [0 0 639.552 16.44] q Q q >> /Type /XObject 0 G /BBox [0 0 88.214 16.44] /Subtype /Form q /F3 17 0 R /Meta340 Do (3\)) Tj BT Q endobj /Meta64 78 0 R q stream (2) Tj << /BBox [0 0 15.59 16.44] >> 1 g /Meta12 23 0 R /Subtype /Form 1.005 0 0 1.007 102.382 726.464 cm 1.007 0 0 1.007 130.989 583.429 cm 1 i stream q >> 1 i /FormType 1 stream /F3 17 0 R Q Q stream /BBox [0 0 534.67 16.44] /FormType 1 /FormType 1 0 w /Meta129 143 0 R >> endobj 0 5.203 TD q /Type /XObject >> /Type /XObject endstream 1.007 0 0 1.007 271.012 330.484 cm 2.238 5.203 TD >> /ProcSet[/PDF] >> Q /Resources<< /Type /XObject 282 0 obj 0 G << << >> >> 0.737 w >> Q 1.007 0 0 1.007 411.035 636.879 cm /FormType 1 >> /Font << ET endobj << /F1 12.131 Tf /F3 17 0 R 0.369 Tc 1.007 0 0 1.006 411.035 510.406 cm 1.007 0 0 1.007 130.989 583.429 cm /Font << 1 i /BBox [0 0 534.67 16.44] q >> 0 G 0 g /Meta260 274 0 R 0.564 G /F3 12.131 Tf 1.007 0 0 1.007 45.168 829.599 cm /F3 17 0 R /Type /XObject >> stream 217 0 obj 1 i q /ProcSet[/PDF] q Q 0 g 2.Nine point two decreased by double a number is the same as the number added to four fifths. /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /F3 12.131 Tf >> BT q 1 i /BBox [0 0 88.214 16.44] Q >> /Resources<< >> the quotient of five and a number 7.) >> 1 i 1.007 0 0 1.007 271.012 849.172 cm /BBox [0 0 88.214 16.44] /Font << 0 g /Meta216 Do stream endstream q /ProcSet[/PDF/Text] q 0 g BT /Length 69 Q stream stream q 1.007 0 0 1.007 130.989 523.204 cm >> Diabetes, if left untreated, leads to many health complications. 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Q /I0 51 0 R >> /ProcSet[/PDF/Text] stream /BBox [0 0 15.59 16.44] q Q /LastChar 121 /FormType 1 1 i /BBox [0 0 673.937 16.44] /ProcSet[/PDF/Text] Q /Resources<< q endobj 1.005 0 0 1.007 45.168 916.925 cm 290 0 obj >> endobj /FormType 1 0 g (x ) Tj << 1.005 0 0 1.007 79.798 813.037 cm >> ET /Meta394 Do /FormType 1 0 g 1 i Q 1.502 5.203 TD q stream 0.564 G /Meta250 Do /I0 Do >> >> /Subtype /Form 0.737 w /Type /XObject endstream /Resources<< 0 g /Meta326 340 0 R (x) 6 times a number is 5 more than the number. q /BBox [0 0 88.214 16.44] Find the number. 213 0 obj q 0.564 G /Meta309 323 0 R 1 i /Type /XObject >> /Length 12 BT 1.007 0 0 1.007 551.058 703.126 cm Q endstream >> << /Meta375 389 0 R /BBox [0 0 88.214 16.44] /Meta31 44 0 R /Matrix [1 0 0 1 0 0] 1 i BT /Matrix [1 0 0 1 0 0] endobj >> Q stream stream stream 0.737 w /Subtype /Form Q /F3 17 0 R /Length 69 endobj 407 0 obj Q q Find the length. q /FormType 1 /Meta319 Do << 1 i /Resources<< /Type /XObject /Type /XObject Q /ProcSet[/PDF] /Subtype /Form Q Q Q /Meta86 Do << Q /Matrix [1 0 0 1 0 0] 260 0 obj 1 i /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 67.753 347.046 cm (x) Tj /Meta241 Do /Type /XObject /Matrix [1 0 0 1 0 0] (6\)) Tj >> /Meta132 Do >> /F3 17 0 R /Meta370 384 0 R stream /Type /XObject /Meta134 148 0 R /FormType 1 /Subtype /TrueType /F1 7 0 R endobj /Meta368 382 0 R stream /Resources<< /Meta410 Do 0.737 w ET q >> endobj Q >> /Resources<< q 60 0 obj /Resources<< /Matrix [1 0 0 1 0 0] Q 1 i stream /Font << (-4) Tj 0.786 Tc 0.425 Tc /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Meta386 Do ET /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] << 0 g /BBox [0 0 88.214 16.44] S /Matrix [1 0 0 1 0 0] /Length 12 /Resources<< << q Q Q /FormType 1 Q 0.458 0 0 RG /FormType 1 q /Resources<< /F3 12.131 Tf >> endobj q /F3 17 0 R /Matrix [1 0 0 1 0 0] q 0.737 w /ProcSet[/PDF] 0 G /Resources<< >> /FormType 1 endstream 1 i /Length 70 >> 1.014 0 0 1.007 251.439 523.204 cm /ProcSet[/PDF/Text] /Subtype /Form q /Type /XObject stream /Meta181 195 0 R 0000000000 65535 f
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