4 years ago

Find two consecutive whole numbers that the square root of 129 lies in between

1 answer:

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Let x and x + 1 (consecutive integers differ by 1) be the two numbers. Use the fact that their sum is equal to 129 to write the equation

x + (x + 1) = 129

x = 64

x = 64 and x + 1 = 65

We can see that the sum of the two numbers is 129

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F(x) = - 2x<br> g(x) = 8x^2– 5x+7<br> Find (f .g)(x).

N76 [4]

Answer:

$\large\boxed{(f\cdot g)(x)=-16x^3+10x^2-14x}$

Step-by-step explanation:

$f(x)=-2x\\\\g(x)=8x^2-5x+7\\\\(f\cdot g)(x)=f(x)\cdot g(x)\\\\\text{substitute:}\\\\(f\cdot g)(x)=(-2x)(8x^2-5x+7)\qquad\text{use the distributive property}\\\\(f\cdot g)(x)=(-2x)(8x^2)+(-2x)(-5x)+(-2x)(7)\\\\(f\cdot g)(x)=-16x^3+10x^2-14x$