Verified Answer :
\(1.\\(4x^2+15x-3)-(-3x^2+5)\\=4x^2+15x-3+3x^2-5\\=7x^2+15x-8\leftarrow A.\)
\(2.\\-3f^2+4f-3+8f^2+7f+1\\=5f^2+11f-2\leftarrow C.\)
\(3.\\(2x^2+6x+1)+(-7x^2+2x-3)\\=2x^2+6x+1-7x^2+2x-3\\=-5x^2+8x-2\leftarrow B.\)
\(4.\\4x^2+3x-3\\\{4;\ 3;-3\}\)
\(5.\\6x^4+3x^3-2x^2+15x-14\\\{6;\ 3;-2;\ 15;-14\}\to5\leftarrow A.\)
\(6.\\-7x-5x^2+5\\\{-7\}\leftarrow D.\)
\(7.\\(2.5\cdot10^4)(4\cdot10^3)=2.5\cdot4\cdot10^{4+3}=10\cdot10^7\\=10^{1+7}=10^8=1\cdot10^8\leftarrow C.\)
\(8.\\2^2\cdot2^8=2^{2+8}=2^{10}\leftarrow B.\)
Answer:
(1) A
(2) C
(3) B
(4) The coefficients are 4,3,-3.
(5) A
(6) D
(7) C
(8) B
Step-by-step explanation:
(1)
The given expression is
\((4x^2 + 15x - 3) - (-3x^2 + 5)\)
Using distributive property.
\((4x^2 + 15x - 3) - (-3x^2) -( 5)\)
\(4x^2 + 15x - 3 + 3x^2 - 5\)
On combining like terms.
\((4x^2+3x^2) + 15x +(- 3 - 5)\)
\(7x^2 + 15x-8\)
Therefore, the correct option is A.
(2)
The given expression is
\(-3f^2 + 4f - 3 + 8f^2 + 7f + 1\)
On combining like terms.
\((-3f^2+ 8f^2) +( 4f + 7f )+(- 3 + 1)\)
\(5f^2 +11f -2\)
Therefore, the correct option is C.
(3)
The given expression is
\((2x^2 + 6x + 1) + (-7x^2 + 2x - 3)\)
Combined like terms.
\((2x^2-7x^2) + (6x+ 2x) + (1 - 3)\)
\(-5x^2 + 8x -2\)
Therefore, the correct option is B.
(4)
The given expression is
\(4x^2 + 3x - 3\)
A number before variable terms are called coefficient of that term.
Therefore, the coefficients are 4,3,-3.
(5)
The given expression is
\(6x^4 + 3x^3 - 2x^2 + 15x - 14\)
It this polynomial, the number of terms is 5.
Therefore the correct option is A.
(6)
The given expression is
\(-7x - 5x^2 + 5\)
The coefficient to x is -7.
Therefore, the correct option is D.
(7)
The given expression is
\((2.5\cdot 10^4)(4\cdot 10^3)\)
\((2.5\cdot 4)\cdot (10^4\cdot 10^3)\)
Using product property of exponent.
\((10)\cdot 10^{4+3}\)
\(1\cdot 10^{1+4+3}\)
\(1\cdot 10^{8}\)
Therefore, the correct option is C.
(8)
The given expression is
\(2^2\cdot 2^8\)
Using product property of exponent.
\(2^{2+8}\)
\(2^{10}\)
Therefore, the correct option is B.