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2 years ago

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Dr. black is standing 15 feet from a streetlamp. the lamp is making his shadow 8 feet long. he estimates the angle of elevation

from the tip of his shadow to the top of the streetlamp is 50 degrees. to the nearest foot how tall is the streetlamp?

1 answer:

ahrayia [7]2 years ago

4 0

Answer:

27 feet

Step-by-step explanation:

Total distance from base of lamp to tip of shadow

15 + 8 = 23

-------------------------

tan∅ = opp/adj

opp will be the height of the lamp

adj is 23

--------------------------

tan 50 = opp/23

multiply both sides by 23

23 * tan50 = opp

Use calculator

27.4 = opp

Rounded

27 feet

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3 years ago

Solve for x<br> 3r – 5r + 10 = 28

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3 years ago

Read 2 more answers

If <img src="https://tex.z-dn.net/?f=x%20%3D%209%20-%204%5Csqrt%7B5%7D" id="TexFormula1" title="x = 9 - 4\sqrt{5}" alt="x = 9 -

Komok [63]

Observe that

$\left(\sqrt x - \dfrac1{\sqrt x}\right)^2 = \left(\sqrt x\right)^2 - 2\sqrt x\dfrac1{\sqrt x} + \left(\dfrac1{\sqrt x}\right)^2 = x - 2 + \dfrac1x$

Now,

$x = 9 - 4\sqrt5 \implies \dfrac1x = \dfrac1{9-4\sqrt5} = \dfrac{9 + 4\sqrt5}{9^2 - \left(4\sqrt5\right)^2} = 9 + 4\sqrt5$

so that

$\left(\sqrt x - \dfrac1{\sqrt x}\right)^2 = (9 - 4\sqrt5) - 2 + (9 + 4\sqrt5) = 16$

$\implies \sqrt x - \dfrac1{\sqrt x} = \pm\sqrt{16} = \pm 4$

To decide which is the correct value, we need to examine the sign of $\sqrt x - \frac1{\sqrt x}$ . It evaluates to 0 if

$\sqrt x = \dfrac1{\sqrt x} \implies x = 1$

We have

$9 - 4\sqrt5 = \sqrt{81} - \sqrt{16\cdot5} = \sqrt{81} - \sqrt{80} > 0$

Also,

$\sqrt{81} - \sqrt{64} = 9 - 8 = 1$

and $\sqrt x$ increases as $x$ increases, which means

$0 < 9 - 4\sqrt5 < 1$

Therefore for all $0 < x < 1$ ,

$\sqrt x - \dfrac1{\sqrt x} < 0$

For example, when $x=\frac14$ , we get

$\sqrt{\dfrac14} - \dfrac1{\sqrt{\frac14}} = \dfrac1{\sqrt4} - \sqrt4 = \dfrac12 - 2 = -\dfrac32 < 0$

Then the target expression has a negative sign at the given value of $x$ :

$x = 9-4\sqrt5 \implies \sqrt x - \dfrac1{\sqrt x} = \boxed{-4}$

Alternatively, we can try simplifying $\sqrt x$ by denesting the radical. Let $a,b,c$ be non-zero integers ( $c>0$ ) such that

$\sqrt{9 - 4\sqrt5} = a + b\sqrt c$

Note that the left side must be positive.

Taking squares on both sides gives

$9 - 4\sqrt5 = a^2 + 2ab\sqrt c + b^2c$

Let $c=5$ and $ab=-2$ . Then

$a^2+5b^2=9 \implies a^2 + 5\left(-\dfrac2a\right)^2 = 9 \\\\ \implies a^2 + \dfrac{20}{a^2} = 9 \\\\ \implies a^4 + 20 = 9a^2 \\\\ \implies a^4 - 9a^2 + 20 = 0 \\\\ \implies (a^2 - 4) (a^2 - 5) = 0 \\\\ \implies a^2 = 4 \text{ or } a^2 = 5$

$a^2 = 4 \implies 5b^2 = 5 \implies b^2 = 1$

$a^2 = 5 \implies 5b^2 = 4 \implies b^2 = \dfrac45$

Only the first case leads to integer coefficients. Since $ab=-2$ , one of $a$ or $b$ must be negative. We have

$a^2 = 4 \implies a = 2 \text{ or } a = -2$

Now if $a=2$ , then $b=-1$ , and

$\sqrt{9 - 4\sqrt5} = 2 - \sqrt5$

However, $\sqrt5 > \sqrt4 = 2$ , so $2-\sqrt5$ is negative, so we don't want this.

Instead, if $a=-2$ , then $b=1$ , and thus

$\sqrt{9 - 4\sqrt5} = -2 + \sqrt5$

Then our target expression evaluates to

$\sqrt x - \dfrac1{\sqrt x} = -2 + \sqrt5 - \dfrac1{-2 + \sqrt5} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - \dfrac{-2 - \sqrt5}{(-2)^2 - \left(\sqrt5\right)^2} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 + \dfrac{2 + \sqrt5}{4 - 5} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - (2 + \sqrt5) = \boxed{-4}$

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1 year ago

Which is less 2/5 or 4/5

KatRina [158]

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$\large\blue\textsf{\textbf{\underline{\underline{Question:-}}}}$

Which is less, 2/5 or 4/5?

$\large\blue\textsf{\textbf{\underline{\underline{Answer and How to Solve:-}}}}$

I'll use a cake for explanation.

Suppose we have a cake, and one of your companions, Mary, is hungry, so she takes 2/5 of one cake. Then, your other friend, Bella, takes 4/5 of the cake.

Then you compare the cakes. Who has less, Mary or Bella?

That's right, Mary.

So 2/5 is less than 4/5.

<h3>Good luck.</h3>

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2 years ago