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3 years ago

Please help, I I beg you please

Mathematics

1 answer:

Tatiana [17]3 years ago

8 0

Answer:

The volume of the cube is $\mathit{\frac{27}{z^3x^{9}}}$ cu in.

Step-by-step explanation:

The Volume of a Cube

Let's have a cube of side length a. The volume of the cube is:

$V=a^3$

The cube of the image has a side length of

$\displaystyle a=\frac{3x^{-3}}{z}\ inches$

Simplifying the expression of the base by converting the negative exponent in the numerator to the denominator:

$\displaystyle a=\frac{3}{zx^{3}}\ inches$

Now find the volume:

$\displaystyle V=\left(\frac{3}{zx^{3}}\ inches\right)^3$

Applying the exponents:

$\displaystyle V=\frac{3^3}{z^3x^{9}}\ inches^3$

$\displaystyle V=\frac{27}{z^3x^{9}}\ inches^3$

The volume of the cube is $\mathbf{\frac{27}{z^3x^{9}}}$ cu in.

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Answer:

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3 years ago

The angle of elevation to a nearby tree from a point on the ground is measured to be 20∘

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Answer:

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Step-by-step explanation:

This is about the right triangle, where 92 feet side is adjacent leg to 20° angle, x - height.

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2 years ago

Read 2 more answers

There are 20 juice cups and each juice cup holds 10 fluid ounces. the juice comes in 1 gallon bottles. how many 1 gallon bottles

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2 bottles

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2 years ago

C=0.41+0.26(z-1) solve for z

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$z = - 15-100c / 26$

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3 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$