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2 years ago

When the angle of elevation of the sun is 78 degrees, a tree casts a 13 foot shadow. How tall is the tree?

Mathematics

1 answer:

oksian1 [2.3K]2 years ago

4 0

Answer: 61.16 ft

Step-by-step explanation:

We can think in this situation as a triangle rectangle.

where:

The height of the tree is one cathetus

The shadow of the tree is the other cathetus.

We know that the angle of elevation of the sun is 78°, an angle of elevation is measured from the ground, then the adjacent cathetus to this angle is the shadow of the tree. And the opposite cathetus will be the height of the tree.

Now we can remember the relationship:

Tg(A) = (opposite cathetus)/(adjacent cathetus)

Where:

A = 78°

Adjacent cathetus = 13ft

opposite cathetus = height of the tree = H

Then we have the equation:

Tg(78°) = H/13ft

Tg(78°)*13ft = H = 61.16 ft

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Find the circumference of a circle with a diameter of 9 in.

Arturiano [62]

Answer:

28.27 inches

Step-by-step explanation:

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2 years ago

1.Arsenic-74 is used to locate brain tumors. It has a half-life of 17.5 days. 90 mg wereused in a procedure. Write an equation t

Sophie [7]

$1)\text{ }N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}$

2) 5.625 mg will be left

Explanation:

1) Half-life = 17.5 days

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To get the equation, we will use the equation of half-life:

$\begin{gathered} N_t\text{ = N}_0(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}} \\ where\text{ N}_t\text{ = amount remaining} \\ N_0\text{ = initial amount} \\ t_{\frac{1}{2}\text{ }}\text{ = half-life} \end{gathered}$ $N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}$

2) we need to find the remaining amount of Arsenic-74 after 70 days

t = 70

$\begin{gathered} N_t=\text{ 90\lparen}\frac{1}{2})^{\frac{70}{17.5}} \\ N_t\text{ = 5.625 mg} \end{gathered}$

So after 70 days, 5.625 mg will be left

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1 year ago

Factorise fully x^3-x

MA_775_DIABLO [31]

Answer:

$x(x+1)(x-1)$

Step-by-step explanation:

We see that the two terms share an x, so we can pull that out:

$x^3-x=x(x^2-1)$

Now notice that we have a difference of squares ( $x^2-1$ ). Given a difference of squares, $a^2-b^2$ , this can always be factored into $(a+b)(a-b)$ . We can use this in this case: $x^2-1=(x+1)(x-1)$ .