If f(x)=7x-3 and g(x)=x^2-4x, find (f+g)(x).

1 answer:

mote1985 [20]2 years ago

8 0

(f+g)(x) is equivalent to f(x) + g(x)

So that would be (7x-3) + (x²-4x) = x² + 7x - 4x - 3 = x² + 3x - 3

Hope this helps.

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Find the volume of a regular hexahedron if one of the diagonals of its faces is 8 root 2 inches.

Naily [24]

The answer is 114sqrt{6} in³

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Diagonal of a cube D is a hypotenuse of a right triangle which other two legs are face diagonal (f) and length of a side (a):
D² = f² + a²

Face diagonal is a hypotenuse of a right triangle which sides are a and a:
f² = a² + a² =2a²

D² = f² + a²
f² = 2a²

D² = 2a² + a² = 3a²
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Volume of a cube with side a is: V = a³
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We have:
D = 8√2 in

$V= (\frac{D}{ \sqrt{3} } )^{3} =( \frac{8 \sqrt{2} }{ \sqrt{3} } )^{3}=( \frac{8 \sqrt{2} * \sqrt{3} }{ \sqrt{3}* \sqrt{3}} )^{3} =( \frac{8 \sqrt{2*3} }{ \sqrt{3*3}} )^{3} =( \frac{8 \sqrt{6} }{ \sqrt{3^{2} }} )^{3} =( \frac{8 \sqrt{6} }{ 3} )^{3} = \\ \\ = \frac{ 8^{3} *( \sqrt{6} )^{3}}{3^{3}} = \frac{512* \sqrt{6 ^{2} }* \sqrt{6} }{27} = \frac{512*6* \sqrt{6} }{27} = 114sqrt{6}$

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Answer:

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Step-by-step explanation:

she had 2 2/12 and she used 1 5/12 so you subtract them and get 9/12

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