twice a number decreased by 58twice a number decreased by 58
ET stream Q 0 g 0 G << /Meta132 Do /Subtype /Form /Subtype /Form (+) Tj stream q /BBox [0 0 88.214 16.44] endobj /ItalicAngle 0 endstream /FormType 1 /BBox [0 0 23.896 16.44] 9 + x. fourteen decreased by a number p. 14 - p. seven less than a number t. t - 7. the product of 9 and a number n. /ProcSet[/PDF/Text] /Meta7 18 0 R /Matrix [1 0 0 1 0 0] Q /Meta248 262 0 R endobj >> q 1 i 16.469 5.336 TD 1 i /Matrix [1 0 0 1 0 0] ET 0.458 0 0 RG /Resources<< >> /FormType 1 ET /F3 17 0 R endstream /F3 12.131 Tf >> /Subtype /Form ET Q /ProcSet[/PDF/Text] /F3 17 0 R stream /Meta175 189 0 R /Meta409 Do q Q endobj [(1)-25(0\))] TJ endstream -0.021 Tw 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 q << 1.014 0 0 1.007 531.485 277.035 cm /Matrix [1 0 0 1 0 0] q 0.737 w Q endstream endobj /ProcSet[/PDF] >> /F4 12.131 Tf /ProcSet[/PDF/Text] Q /MaxWidth 1248 Q << /Meta409 425 0 R /Matrix [1 0 0 1 0 0] stream 0 G /Meta166 Do q endobj /Subtype /Form /Matrix [1 0 0 1 0 0] stream q 1 i 385 0 obj /F3 17 0 R Q >> 0 G BT /Matrix [1 0 0 1 0 0] /Type /XObject /Resources<< 0 G /Meta37 50 0 R Q endobj Q 1 i endstream q /Meta231 Do /ProcSet[/PDF/Text] q << /FormType 1 (-23) Tj /FontName /TestGen-Regular q 1.007 0 0 1.007 271.012 776.149 cm /Meta208 222 0 R /BBox [0 0 88.214 35.886] 0 g Q /Resources<< ET >> >> /Length 69 /FormType 1 endstream endstream stream /Font << 0.458 0 0 RG w/Honors. stream >> q q endstream /Resources<< /Info 3 0 R 1 i /F3 17 0 R /F3 17 0 R Q stream /ProcSet[/PDF/Text] q BT endstream 1 i /F3 12.131 Tf ET 1 i 0.458 0 0 RG 12.727 5.203 TD << /FormType 1 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. 1.007 0 0 1.007 411.035 383.934 cm 1 i 1 i /Meta193 207 0 R 672.261 546.541 m 1.007 0 0 1.007 271.012 636.879 cm endstream /Meta22 33 0 R /ProcSet[/PDF/Text] 1 i q Q >> 1 i 1.014 0 0 1.006 531.485 690.329 cm 1.005 0 0 1.007 102.382 363.608 cm /ProcSet[/PDF/Text] Q /Font << q /Subtype /Form /Subtype /Form /StemH 94 q 17.234 5.203 TD /Meta151 Do 95 0 obj Q (x) Tj /Type /XObject 0 G >> 20.21 5.203 TD endobj /Meta251 Do /ProcSet[/PDF/Text] 0.564 G >> /F4 36 0 R 1.007 0 0 1.007 271.012 636.879 cm << q 1 i 1.014 0 0 1.007 531.485 450.181 cm /Subtype /Form endstream 0.369 Tc Q q 0 g Q q 1 i BT q q endobj 0.458 0 0 RG 0.68 Tc Q /Resources<< /BBox [0 0 88.214 16.44] 1 i /Resources<< >> /ProcSet[/PDF/Text] << stream Q 0 G >> /Meta153 167 0 R endobj << /Resources<< Q q /Type /XObject /ProcSet[/PDF/Text] Was this answer helpful? 0 w endstream /FormType 1 >> << 1 i endstream /Font << 1.007 0 0 1.007 130.989 383.934 cm q 1 i 549.694 0 0 16.469 0 -0.0283 cm q 0.369 Tc >> >> Q 110 0 obj /Length 78 /ProcSet[/PDF] q /Type /XObject 0 g /BBox [0 0 639.552 16.44] ET 1 i >> /Meta75 Do 1.007 0 0 1.007 411.035 330.484 cm /Meta52 66 0 R q 279 0 obj 0 G Q << 216 0 obj q 0 g 0 g endobj 1 g /Resources<< endobj 250 0 obj 0.737 w /Resources<< Q Q 0 g 1 i 6.746 5.203 TD 0.738 Tc 399 0 obj 1.007 0 0 1.007 271.012 383.934 cm endobj << /F4 36 0 R /FormType 1 >> BT >> endobj << >> (2) Tj /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] ET 1 g q Q /Subtype /Form 0 g /F3 17 0 R q << /ProcSet[/PDF/Text] /FormType 1 BT >> /F3 17 0 R /BBox [0 0 88.214 16.44] /F3 17 0 R /ProcSet[/PDF/Text] 1 i Q q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 551.058 277.035 cm /Matrix [1 0 0 1 0 0] q /Meta243 257 0 R 0.737 w (\)) Tj /Subtype /Form Q q 431 0 obj stream q endstream q q /F3 12.131 Tf 0.458 0 0 RG /Meta151 165 0 R >> >> /F1 12.131 Tf Medium /ProcSet[/PDF] /Length 54 stream Q /Type /XObject /Meta64 Do Just type into the box and your calculation will happen automatically. 0.458 0 0 RG /Meta103 Do q /Subtype /Form Q /F3 17 0 R 2x - 15 = -27. ET Q >> >> endobj q /Matrix [1 0 0 1 0 0] 20.975 5.336 TD a and b or something else.***. /F3 17 0 R Q /BBox [0 0 88.214 16.44] endstream endobj (x ) Tj 30 0 obj endobj Q stream (-) Tj q /Meta299 313 0 R /BBox [0 0 15.59 16.44] /Length 69 endstream 19.474 5.203 TD (B) Tj /Meta384 Do /Subtype /Form /Type /XObject There were x cookies at the beginning of a party. Q << q Q Q /Parent 1 0 R q In other terms, 52-nx The problem is asking that you subtract twice a number from 52. ET /Meta378 Do /F3 12.131 Tf q stream /Length 12 /F3 17 0 R /Type /XObject 4 less than some number : x - 4 : a number decreased by 10 : y - 10 : 8 minus some number : 8 - t : the difference between a number and 12 : . endstream /Resources<< 1 i /FormType 1 /Matrix [1 0 0 1 0 0] q 1 g stream /BBox [0 0 30.642 16.44] 1.014 0 0 1.006 251.439 763.351 cm /Length 69 ET /Type /XObject /Resources<< /F3 17 0 R q q /BBox [0 0 88.214 16.44] 0.458 0 0 RG /Type /XObject /ProcSet[/PDF] /Meta104 Do /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 551.058 383.934 cm ET /BBox [0 0 88.214 16.44] /Type /XObject /Matrix [1 0 0 1 0 0] 140781 /Length 118 >> endstream endstream ET q 0 g 52 0 obj (+) Tj >> Q /FormType 1 >> 1.007 0 0 1.006 551.058 763.351 cm 0 g 0.369 Tc >> ET /Meta176 Do 0.271 Tc q /Subtype /Form stream 1 i /BBox [0 0 15.59 16.44] >> 0.564 G << /Matrix [1 0 0 1 0 0] Q /Meta48 Do 1 i endstream q Q >> 0.564 G /F3 12.131 Tf /Font << Q >> Q 439 0 obj ( x) Tj 102 0 obj 341 0 obj /BBox [0 0 534.67 16.44] 445 0 obj 1.007 0 0 1.006 130.989 437.384 cm 1 i /Subtype /Form /F3 12.131 Tf /ProcSet[/PDF/Text] Q >> 0 4.894 TD 261 0 obj 0 g << >> /Length 68 q /Font << /F3 12.131 Tf 1.007 0 0 1.007 551.058 703.126 cm /FormType 1 1.014 0 0 1.007 111.416 330.484 cm If LtitnS6S . /F3 12.131 Tf 208 0 obj >> /ProcSet[/PDF/Text] Q Q 0 G << endstream 11.99 24.649 TD /ProcSet[/PDF/Text] 0 g /Font << /Meta338 352 0 R q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf >> 271 0 obj 336 0 obj ET /Resources<< << 0 G /F3 12.131 Tf q /ProcSet[/PDF/Text] /Length 59 /Meta65 79 0 R BT /Resources<< For the lesson, he grabs a glass container shaped like a rectan /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] 0 g endobj BT endobj 1 i >> /BBox [0 0 17.177 16.44] -0.101 Tw Q (x ) Tj Q /Matrix [1 0 0 1 0 0] 2x - y = 6. x + 3y = -25. 1.005 0 0 1.007 45.168 889.071 cm q >> /BBox [0 0 88.214 16.44] q q endstream >> /Meta231 245 0 R >> /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 251.439 583.429 cm stream 43 0 obj /BBox [0 0 17.177 16.44] 0.737 w /ProcSet[/PDF/Text] Q /Length 206 >> q /Length 69 Q Q endobj /MissingWidth 250 q /Resources<< 1 i endobj (5) Tj /FormType 1 << BT /F3 12.131 Tf q Q Q 0 w >> /Subtype /Form Q stream /Meta307 321 0 R /Font << << /Meta237 251 0 R /Length 65 ET (40) Tj 0 w >> << /ProcSet[/PDF/Text] << q /F1 12.131 Tf Q Q Q /Length 69 q Q >> q /Length 69 endobj endobj /Type /XObject /Length 245 0 w /Font << endstream /Meta148 162 0 R /Length 59 q /Type /XObject /Matrix [1 0 0 1 0 0] 328 0 obj Q q 1.007 0 0 1.007 411.035 849.172 cm >> (8\)) Tj q 0.134 Tc /Length 79 /Leading 349 >> ET >> endstream 0 g /FormType 1 /Meta10 21 0 R 0.564 G /Length 2252 1.007 0 0 1.007 130.989 636.879 cm 53 0 obj Q 446 0 obj q endobj Find the number. >> endobj >> /Meta55 Do >> endobj Q q q q endobj q 1 i q q 405 0 obj /Length 69 281 0 obj << Q /Meta353 367 0 R /Subtype /Form /Resources<< BT 0 G /Meta48 62 0 R >> endstream /Font << /Length 54 /Matrix [1 0 0 1 0 0] /Length 12 /ProcSet[/PDF/Text] /Meta76 90 0 R Q 0.737 w stream q 0.458 0 0 RG Q /ProcSet[/PDF] Q /Matrix [1 0 0 1 0 0] Q 0 g 0.564 G /Resources<< 1.014 0 0 1.007 531.485 523.204 cm endstream 0 g 1 i /Meta320 Do /ProcSet[/PDF] stream 0 g 0 g ET /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF] >> 0.737 w /Font << /Meta1 8 0 R Q /F3 17 0 R /Type /XObject /Meta115 129 0 R 0.369 Tc >> << 0.737 w endstream /Length 16 q 3.742 5.203 TD 1 i >> ET ET q 1 i endstream /F3 17 0 R Q Q /ProcSet[/PDF/Text] endstream /Matrix [1 0 0 1 0 0] (D\)) Tj ET Q >> 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 /Resources<< 258 0 obj >> /Length 69 0 G 0 g 0 G 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. /ProcSet[/PDF] /Meta263 277 0 R /F3 12.131 Tf /Type /XObject /Subtype /Form /Length 65 /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Font << 1 g stream 0.564 G /Resources<< /F3 17 0 R 0.564 G /Length 16 q 0.425 Tc endstream >> << /Meta285 Do >> >> [(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 >> 0 g /Length 69 /FormType 1 /BBox [0 0 639.552 16.44] 0.458 0 0 RG /Resources<< 1 i >> q /Font << >> /Resources<< (-11) Tj /Meta360 Do (A\)) Tj 0.155 Tc /FormType 1 /Meta296 Do >> << /F3 12.131 Tf endstream 0 w /ProcSet[/PDF/Text] /I0 51 0 R >> >> >> q /StemH 88 ET 1.007 0 0 1.007 411.035 330.484 cm /Meta301 Do q Q << (D\)) Tj 1.007 0 0 1.007 411.035 583.429 cm q 1 i [(The )-19(quotient of )] TJ /Length 12 0 G endobj /F3 17 0 R /BBox [0 0 88.214 35.886] 0 g q q /F4 12.131 Tf >> >> endobj /FormType 1 ET /Subtype /Form /ProcSet[/PDF/Text] >> /Type /XObject >> /ProcSet[/PDF/Text] /Resources<< /Type /XObject endstream 317 0 obj Q /F3 12.131 Tf 35.206 4.894 TD /F3 17 0 R q /Subtype /Form q /Meta268 282 0 R BT /Meta276 290 0 R /FormType 1 /F1 12.131 Tf 0.458 0 0 RG 0 g stream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] [(MULTIPLE CHOICE. (38) Tj endobj /Type /XObject endobj 0 G 0 g 0 5.336 TD /XHeight 471 /Meta229 243 0 R 1 i 20.21 5.203 TD 672.261 726.464 m /FormType 1 endobj Q /Descent -299 stream /FormType 1 q >> 1.014 0 0 1.007 391.462 636.879 cm q q 323 0 obj We determined the effect of plant oils (rapeseed, sunflower, linseed) and organic acids (aspartic and malic) on the fermentation of diet consisting of hay, barley and sugar beet molasses. /Resources<< /Length 16 /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1 i endobj Q /F3 17 0 R /Meta74 Do endstream ET /ProcSet[/PDF] . q >> /Meta168 182 0 R 1.005 0 0 1.007 102.382 400.496 cm 0 w /FormType 1 q 1 i /Meta2 Do 1.007 0 0 1.007 411.035 636.879 cm /Length 16 -0.486 Tw Q /Meta221 Do >> << endstream 549.694 0 0 16.469 0 -0.0283 cm endobj /BBox [0 0 549.552 16.44] stream /Subtype /Form 1 g 26.957 5.203 TD /Resources<< 1 i >> q 1.502 5.203 TD endstream Q /Meta299 Do endstream Q /Matrix [1 0 0 1 0 0] << 1 i 0.458 0 0 RG Twice a number is decreased by 9, and this sum is multiplied by 4. endobj Q Q ET /FormType 1 /Type /XObject 0 g >> >> >> >> q /Resources<< Q << /BBox [0 0 15.59 16.44] q /Type /XObject /Meta179 Do /FormType 1 /ProcSet[/PDF/Text] 1.007 0 0 1.007 271.012 277.035 cm /F3 12.131 Tf /FormType 1 /BBox [0 0 534.67 16.44] Q /BBox [0 0 15.59 16.44] 354 0 obj Q 30.699 5.203 TD 0.564 G /Matrix [1 0 0 1 0 0] 0 G /ColorSpace [/Indexed /DeviceGray 1 ] /FormType 1 /BBox [0 0 534.67 16.44] /Type /XObject /Resources<< /ProcSet[/PDF] 0.369 Tc /Subtype /Form >> >> Q 0.37 Tc Q /Meta411 427 0 R 3.742 5.203 TD ET Q /Font << Q q /Resources<< /Matrix [1 0 0 1 0 0] << endobj /F3 17 0 R ET /F3 17 0 R endobj /F3 12.131 Tf >> /Length 16 /Matrix [1 0 0 1 0 0] endobj /BBox [0 0 534.67 16.44] q /Length 70 420 0 obj Q /Subtype /Form q 193 0 obj 0 g Q /FormType 1 endstream Answer link. 380 0 obj q endobj 1.014 0 0 1.006 251.439 510.406 cm q stream 1.007 0 0 1.007 411.035 330.484 cm q /Type /XObject 0 g << /Length 69 << 0 5.203 TD Q 0.37 Tc Results: patients with type 2 diabetes mellitus predominated with 95.0 %. Q Q /ProcSet[/PDF/Text] /Meta278 292 0 R Q 432 0 obj q Find the number. stream 211 0 obj BT stream 0.68 Tc /Subtype /Form /Resources<< q 359 0 obj q >> << 0.564 G /Meta142 Do >> 0 g ET endstream 0.737 w /Meta218 232 0 R /Meta45 59 0 R /Length 69 endobj /Length 16 /Meta80 94 0 R 0.738 Tc >> /FormType 1 /Matrix [1 0 0 1 0 0] stream << 333.269 5.488 TD /Subtype /Form /FormType 1 0 5.203 TD endobj BT (5) Tj Q /Subtype /Form /Length 57 "49 . 0.738 Tc (7\)) Tj >> /Subtype /Form q /F3 12.131 Tf /XObject << endobj q 0.458 0 0 RG Q /FormType 1 >> >> 722.699 546.541 l /Meta118 Do /BBox [0 0 15.59 16.44] /FormType 1 >> /Meta187 201 0 R >> stream /Subtype /Form << /Meta252 Do endstream 0 g endstream 0 G /F3 17 0 R endobj 0.458 0 0 RG stream /BBox [0 0 639.552 16.44] [( and )16(a nu)26(mbe)18(r)] TJ >> A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. That was 1/8 of the points that he scored q /Resources<< Q endstream q stream /Resources<< /ProcSet[/PDF/Text] stream /ProcSet[/PDF] /Meta62 Do q << /ProcSet[/PDF] >> /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 654.946 473.519 cm /BBox [0 0 534.67 16.44] /Resources<< /F2 11 0 R >> 0.737 w /Length 63 /ProcSet[/PDF/Text] >> 0 g 1.005 0 0 1.007 102.382 293.596 cm endobj << Q /Resources<< Q endobj 0 g Q /Matrix [1 0 0 1 0 0] /FormType 1 /FormType 1 0 G (5) Tj /Meta246 Do >> q >> 0.564 G /Length 70 /Meta302 316 0 R 1 i q endstream 1 i 1 i ET /Resources<< Q q You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8 gives 58 find the number, 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. 1.014 0 0 1.007 531.485 330.484 cm ET /F3 12.131 Tf 1 g /Type /XObject << ET BT /Length 69 endstream 0.564 G BT ET /Resources<< 1.007 0 0 1.006 411.035 437.384 cm 152 0 obj >> /Resources<< 1.007 0 0 1.007 271.012 583.429 cm /Subtype /TrueType Q /Subtype /Form /Subtype /Form Q (\)) Tj 0 5.203 TD Q 0 G /Matrix [1 0 0 1 0 0] /F3 12.131 Tf >> /Subtype /Form /Matrix [1 0 0 1 0 0] /Type /XObject Q Q 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 Q /Resources<< 1 i /FormType 1 /Meta404 420 0 R /Resources<< /Length 69 /Subtype /Form /Type /XObject /F3 17 0 R 119 0 obj >> 1.007 0 0 1.007 271.012 383.934 cm << /Type /XObject >> /Matrix [1 0 0 1 0 0] Q 0 5.203 TD BT /Length 69 q /BBox [0 0 639.552 16.44] Q stream Q /FormType 1 9.723 5.336 TD /Meta76 Do /Type /FontDescriptor endobj /Length 59 1.007 0 0 1.007 411.035 849.172 cm q stream /Resources<< 0.458 0 0 RG /F3 12.131 Tf BT /FormType 1 q q q (5\)) Tj /Meta416 Do 1 i /Resources<< 1 i q -0.056 Tw /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 654.946 799.486 cm 267 0 obj 0 G 0 g endobj /Subtype /Form BT /Length 60 >> /Meta22 Do 0 g /Type /XObject /Matrix [1 0 0 1 0 0] /Meta197 211 0 R S << endobj /Meta264 278 0 R 549.694 0 0 16.469 0 -0.0283 cm /FormType 1 (D\)) Tj /Subtype /Form Q /Type /XObject >> 0.51 Tc 1 i Q >> Q q Q /Subtype /Form 0 g (x) Tj /Subtype /Form 292 0 obj /Type /XObject 20.21 5.203 TD /Length 69 /ProcSet[/PDF/Text] (58) Tj /F3 12.131 Tf /Subtype /Form 0.458 0 0 RG >> endobj /Meta68 82 0 R /Meta417 Do /Font << 108 0 obj endstream stream /Subtype /Form q (B\)) Tj /F1 12.131 Tf stream /Subtype /Form 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 /Type /XObject (1\)) Tj /Type /XObject /F3 17 0 R /FormType 1 /F3 12.131 Tf << /XObject << >> endobj /Length 104 The sum of 18 and tour times a number is -6 Find the number. q q Q 1.007 0 0 1.007 271.012 450.181 cm /Type /XObject 344 0 obj 1.007 0 0 1.007 271.012 636.879 cm stream endstream 0 g /Meta15 26 0 R >> /Length 54 /Resources<< 0 w stream /Meta381 395 0 R q 242 0 obj endstream /F3 12.131 Tf >> q 0 5.203 TD (x) 6 times a number is 5 more than the number. << 0 G /Subtype /Form /Resources<< >> q Q q endobj BT q /I0 Do 194 0 obj q /F3 17 0 R 1 i Q /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] q 333.269 5.488 TD Q endstream /Resources<< q ET Q /Type /XObject /Subtype /Form endstream q /Resources<< /FormType 1 q /FormType 1 Q Q BT endobj q << Q endobj (- 8) Tj /Length 70 The first number increased by three times the second number is -25. x + 3y = -25. by solving the system of equations. /Resources<< /Resources<< /Subtype /Form endobj q q /Meta146 160 0 R /Length 73 q 0.737 w /Matrix [1 0 0 1 0 0] /Type /Font Q << /Meta141 155 0 R Q 0.786 Tc q /Font << /FormType 1 1 g /Resources<< 0 G /BBox [0 0 88.214 35.886] 0 g 0 g 1 i endstream /ProcSet[/PDF] /Meta219 Do /Matrix [1 0 0 1 0 0] q endobj stream /Resources<< /Subtype /Form [tex]\sin (\pi -x)=\sin x[/tex]. BT /Meta422 Do endobj 263 0 obj 0 w 6.746 5.203 TD 0 G /Type /XObject stream >> Q (-) Tj >> q 71 0 obj endstream Q /BBox [0 0 17.177 16.44] 0 g endobj -0.16 Tw Making educational experiences better for everyone. /FormType 1 Q /ProcSet[/PDF/Text] /Font << (-8) Tj << ET Q 0.458 0 0 RG 0 g Q >> ET q endobj >> Q q endobj /Resources<< q Q 1.502 24.339 TD endstream /F3 12.131 Tf 148 0 obj 1.007 0 0 1.007 45.168 779.913 cm /Meta85 Do stream /Meta41 55 0 R BT [tex]\sin (\pi -x)=\sin x[/tex]. /BBox [0 0 673.937 15.562] /BBox [0 0 88.214 16.44] /Subtype /Form << 0 g /Meta255 269 0 R q stream q /ProcSet[/PDF/Text] /Resources<< /Descent -299 BT /Length 12 q /Type /XObject /Type /XObject Twice = two times, double. endstream /ProcSet[/PDF/Text] /Resources<< 1.014 0 0 1.007 531.485 703.126 cm 0 G endstream q /Matrix [1 0 0 1 0 0] stream /FormType 1 Q 1 i 0.564 G q q /Type /XObject /Subtype /Form /F1 7 0 R /Type /Page /Font << q stream 413 0 obj q /Type /XObject Q /BBox [0 0 15.59 16.44] 1.005 0 0 1.007 79.798 846.161 cm 0.737 w &K @ 0.458 0 0 RG endstream q 0 g BT q /Meta182 Do /Resources<< 225 0 obj 0.458 0 0 RG Andrew M. /Type /XObject /Resources<< BT >> endstream BT /FormType 1 << /BBox [0 0 23.896 16.44] Q Q /BBox [0 0 88.214 16.44] 0 G 1 i Q /ProcSet[/PDF] Q q /F1 12.131 Tf Q Q 0 g Q 1 g Q Q stream Q endstream /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 872.509 cm 1.005 0 0 1.007 102.382 653.441 cm >> Q << << q Q Q 0.28 Tc ( \() Tj Q 0 g Q /Length 69 >> q 167 0 obj /Type /XObject /F3 17 0 R /Resources<< /FormType 1 >> (+) Tj BT 0 g q endstream /F3 17 0 R q Q endobj 0 G /Font << ET 0.737 w /Type /XObject Q 0 5.203 TD /Subtype /Form /Matrix [1 0 0 1 0 0] >> /Meta116 130 0 R /Matrix [1 0 0 1 0 0] /FontName /TimesNewRomanPSMT >> q /Resources<< /ProcSet[/PDF] /Meta307 Do endstream /Meta45 Do << endobj 0 5.203 TD /F3 17 0 R 1.014 0 0 1.007 391.462 330.484 cm /Length 16 Q /Matrix [1 0 0 1 0 0] Q << endobj /Matrix [1 0 0 1 0 0] /Meta337 351 0 R /Meta379 393 0 R You can specify conditions of storing and accessing cookies in your browser, Twice a number, decreased by 58 is less than 112, Mr. Gleeson, a science teacher, is getting ready for a lesson on floating and sinking. Q 1 i q 1.007 0 0 1.007 551.058 636.879 cm /F3 12.131 Tf stream 1 g /F3 12.131 Tf /Meta428 444 0 R 1 i 1.014 0 0 1.006 391.462 690.329 cm /ProcSet[/PDF] /Meta379 Do /Resources<< 1 g /Resources<< Q /Type /XObject /BBox [0 0 17.177 16.44] q /Font << >> /Subtype /Form 0 g 1.005 0 0 1.007 79.798 730.228 cm /Meta249 263 0 R Q 1.007 0 0 1.007 45.168 829.599 cm q q 0 G 0.524 Tc 0 g /ProcSet[/PDF/Text] /ProcSet[/PDF] /Font << >> 16.469 5.336 TD 1 i precision and actual right or wrong answers. /BBox [0 0 88.214 35.886] >> Q << >> 1 i /ProcSet[/PDF/Text] 1.005 0 0 1.007 102.382 347.046 cm /Resources<< stream 0 g q << 188 0 obj /Length 65 /BBox [0 0 88.214 16.44] 342 0 obj /Length 54 /Font << << >> Q BT /Meta362 Do stream stream >> 1.007 0 0 1.007 130.989 636.879 cm /Subtype /Form /Meta350 364 0 R >> - 9737014. << /F3 17 0 R q endstream /F1 7 0 R /Meta27 Do >> 22.478 5.336 TD (40) Tj /FormType 1 /BBox [0 0 534.67 16.44] /Subtype /Form << %PDF-1.4 >> /ProcSet[/PDF] endobj /ProcSet[/PDF] >> q /Matrix [1 0 0 1 0 0] >> /Length 79 >> q /Length 54 >> /Matrix [1 0 0 1 0 0] Q If twice a number is decreased by 13, the result is 9. q >> ET q 442 0 obj /F3 12.131 Tf q stream /Matrix [1 0 0 1 0 0] 0.369 Tc /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] /F3 12.131 Tf << q endstream /Matrix [1 0 0 1 0 0] Q >> 14 0 obj 0 G q /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 654.946 347.046 cm /BBox [0 0 15.59 16.44] q /Type /XObject endobj /Subtype /Form Q 0.564 G 1 g 1.005 0 0 1.007 79.798 829.599 cm >> 1.007 0 0 1.007 271.012 776.149 cm << >> >> 0.737 w /Length 59 q 0 G Q >> /Resources<< ET 0.458 0 0 RG /Meta17 Do 1 i 0 g >> /Font << 5.98 7.841 TD Q 0 G endstream /Resources<< /Meta5 Do /Subtype /Form /FormType 1 endstream 1 i Q 1 g /Matrix [1 0 0 1 0 0] q /Font << /F3 12.131 Tf 1 i Q A) 5 more than a number. Q 1 i /Resources<< >> endobj Q /Meta83 Do stream /Font << /Subtype /Form q BT >> 440 0 obj ET q /Resources<< BT 0 4.894 TD /AvgWidth 445 /F4 36 0 R The sum Of twice a nu4ber What is the number? q /Matrix [1 0 0 1 0 0] stream /Resources<< q 0.425 Tc endstream /Meta14 Do 1 i (vi) If 12 is subtracted from a number, the result is 24. 2.238 5.203 TD q q /Meta353 Do /Meta356 370 0 R Q >> 1 i Q 57.656 5.203 TD BT endobj Q 1 g q (x ) Tj 347 0 obj /Resources<< /Length 69 /Type /XObject q /Type /XObject /Subtype /Form endstream endstream S /Length 16 Q q Q << BT endstream /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] 1.014 0 0 1.006 531.485 836.374 cm 1 i q Three times a number equals fifteen 3. 234 0 obj q /Subtype /Form << >> 272 0 obj /Length 59 1.007 0 0 1.007 411.035 277.035 cm 0 G ET /Meta176 190 0 R q 199 0 obj /Type /XObject 111 0 obj 0 5.203 TD << >> 0 g >> endobj q ET Q >> 319 0 obj /I0 Do ET << >> Q >> >> Q q /Length 64 stream Q endstream >> /F1 12.131 Tf >> BT /Type /XObject Q 1 i 1.007 0 0 1.007 130.989 523.204 cm stream stream /Type /XObject endstream 9.723 5.336 TD /Meta136 150 0 R Twice a number when decreased by 7 gives 45. endstream /Matrix [1 0 0 1 0 0] 1.007 0 0 1.006 411.035 437.384 cm endstream /Type /XObject 1.007 0 0 1.007 411.035 636.879 cm endstream 0000000000 65535 f
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Reynolds Platinum Membership, Melbourne 1980s Restaurants, Articles T
Reynolds Platinum Membership, Melbourne 1980s Restaurants, Articles T