61. How much is 21a 3 – 17a 2 less than 89a 3 – 64a 2 + 6a + 16?
62. How much is y 4 – 12y 2 + y + 14 greater than 17y 3 + 34y 2 – 51y + 68?
63. How much does 93p 2 – 55p + 4 exceed 13p 3 – 5p 2 + 17p – 90?
64. To what expression must 99x 3 – 33x 2 – 13x – 41 be added to make the sum zero?
(From Class 7 NCERT Exemplar Chapter Algebraic Expressions.)
Please help, I will mark the brainliest.
Answers
Answer:
Let the given polynomial is less than the other by M
According to question,
21a
3
−17a
2
+M=89a
3
−64a
2
+6a+16
⇒M=(89a
3
−64a
2
+6a+16)−(21a
3
−17a
2
)
⇒M=89a
3
−64a
2
+6a+16−21a
3
+17a
2
Collecting like terms,
⇒M=89a
3
−21a
3
−64a
2
+17a
2
+6a+16
⇒M=68a
3
−47a
2
+6a+16
Answer:
61. in the photo.
62.y4 -12y2 +y+14-(17y3+34y2 -51y+68) = y4 -12y2 +y+14-17y3-34y2 +51y-68 On combining the like terms, = y4-12y2-34y2 + y+51y+14-68-17y3 = y4-46y2 + 52y-17y3 -54 = y4 -17y3 -46y2 + 52y-54 So, y4 -12y2 +y+14 is y4-17y3-46y2 + 52y-54 greater than 17y3+34y2 -51y+68
63.93p2 -55p+4 —(13p3 -5p2 + 17p-90) = 93p2 -55p+4 -13p3+5p2 – 17p + 90 On combining the like terms, = 93p2 + 5p2-55p-17p+4+90-13p3 = 98p2-72p+94-13p3 =-13p3 + 98p2-72p+ 94 So, 93p2 -55p+4 is -13p3 + 9p2 -72p+ 94 exceed from 13p3 -5p2 + 17p-90.
64.In order to find the solution, we will subtract 99x3 -33x2 -13x-41 from 0. Required expression is 0- (99x3 -33x2 -13x-41) = 0- 99x3 +33x2 +13x+41 = – 99x3 +33x2 +13x+41 So, If we add -99x3 +33x2 +13x+41 to 99x3 -33x2 -13x-41, then the sum is zero.
