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Irina-Kira [14]

3 years ago

15

The statue of Liberty stands on a 150 ft. pedestal. From a point 250 ft. from the base of the

1 answer:

bekas [8.4K]3 years ago

4 0

Answer:

148ft

Step-by-step explanation:

To solve this question, you'll have to imagine the statue makes a right angle triangle with the base since it has an angle of elevation from the base to the top of the torch.

Assuming the height from the pedestal to the top of the torch is y

The height of the statue is x

But we know the height of the pedestal = 150ft

The distance from the observer to the base of the pedestal = 250ft

And the angle of elevation = 50°

See attached document for better illustration.

Tanθ = opposite / adjacent

θ = 50°

Adjacent = 250

Opposite = y

Tan50 = t / 250

y = 50 × tan50

y = 50 × tan50

y = 50 × 1.1917

y = 297.925ft

The height of the statue from the base of the pedestal to the top of the torch is 297.925ft

The height of the statue = x

x = (height of the statue + height of the pedestal) - height of the pedestal

x = y - 150

x = 297.925 - 150

x = 147.925ft

Approximately 148ft

The height of the statue is 148ft

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2 years ago

Given the function: f(x) = 2x3 - 3/2x, find f(-4).

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Step-by-step explanation:

Given:

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$f(-4)= 2\times(-4)^3-\frac{3}{2}\times(-4)\\$

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So we get;

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steposvetlana [31]

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3 years ago

Х- а<br>x-b<br>If f(x) = b.x-a÷b-a + a.x-b÷a - b<br>Prove that: f (a) + f(b) = f (a + b)

GenaCL600 [577]

Given:

Consider the given function:

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}$

To prove:

$f(a)+f(b)=f(a+b)$

Solution:

We have,

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}$

Substituting $x=a$ , we get

$f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}$

$f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}$

$f(a)=0+a$

$f(a)=a$

Substituting $x=b$ , we get

$f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}$

$f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}$

$f(b)=b+0$

$f(b)=b$

Substituting $x=a+b$ , we get

$f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}$

$f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}$

$f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}$

$f(a+b)=\dfrac{b^2-a^2}{b-a}$

Using the algebraic formula, we get

$f(a+b)=\dfrac{(b-a)(b+a)}{b-a}$ $[\because b^2-a^2=(b-a)(b+a)]$

$f(a+b)=b+a$

$f(a+b)=a+b$ [Commutative property of addition]

Now,

$LHS=f(a)+f(b)$

$LHS=a+b$

$LHS=f(a+b)$

$LHS=RHS$

Hence proved.

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2 years ago