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2 years ago

14

A school flag poll casts a 16 foot shadow on the lawn. A teacher stood at the shadows edge and measure the angle of elevation to

the top of the poll at 62°. How tall is the poll?

1 answer:

vovikov84 [41]2 years ago

6 0

Answer:

30.1 feet.

Step-by-step explanation:

We have been given that a school flag poll casts a 16 foot shadow on the lawn. A teacher stood at the shadows edge and measure the angle of elevation to the top of the poll at 62°. We are asked to find the height of the tall.

First of all, we will draw a diagram to represent our give scenario as shown in the attachment.

We can see that side length with 16 feet is adjacent side and height of the pole (h) is opposite side to angle of elevation.

We know that tangent relates opposite side of a right triangle to its adjacent side.

$\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}$

$\text{tan}(62^{\circ})=\frac{h}{16}$

$h=16\cdot \text{tan}(62^{\circ})$

$h=16\cdot 1.880726465346$

$h=30.0916234\approx 30.1$

Therefore, the poll is approximately 30.1 feet tall.

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lord [1]

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3 years ago

Between the years of 1947 and 1956, earthenware jars containing what are known as the Dead Sea Scrolls were found in caves along

lidiya [134]

Answer:

They are right, the time calculated using the exponential decay equation is 1948.6 years.

Step-by-step explanation:

The time can be calculated using the exponential decay equation:

$N_{(t)} = N_{0}e^{-\lambda t}$ (1)

<u>Where</u>:

N(t): is the quantity of C-14 at time t

N₀: is the initial quantity of C-14

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The decay constant is:

$\lambda = \frac{ln(2)}{t_{1/2}}$ (2)

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$t = \frac{t_{1/2}*ln(N_{t}/N_{0})}{ln(2)} = \frac{5730*ln(0.79)}{ln(2)} = 1948.6 y$

Therefore, the time of the scrolls estimated by the archeologists is right, since the time calculated using the exponential decay equation is 1948.6 years.

I hope it helps you!

8 0

2 years ago

I need help with number 1

Sophie [7]

Changing those words into an equation, we get:
7+3x = 19
Subtract both sides by 7
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That number is 4. Have a nice day! :)

6 0

3 years ago

g An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single b

Fed [463]

Answer:

Therefore the of blue in the second urn is 4.

Step-by-step explanation:

Let second urn contain x number of blue ball .

Urn Red Ball Blue Ball Total Ball

1 4 6 10

2 16 x 16+x

Getting a red ball from first urn is $P(R_1)=\frac{\textrm {Number of red ball}}{\textrm {Total ball}}$ $=\frac{4}{10}$

Getting a blue ball from first urn is $P(B_1)=\frac{\textrm {Number of blue ball}}{\textrm {Total ball}}$ $=\frac {6}{10}$

Getting a red ball from second urn is $P(R_2)=\frac{\textrm {Number of red ball}}{\textrm {Total ball}}$ $=\frac{16}{16+x}$

Getting a blue ball from second urn is $P(B_2)=\frac{\textrm {Number of blue ball}}{\textrm {Total ball}}$ $=\frac {x}{16+x}$

Getting two red balls from first and second urn is $=\frac{4}{10}\times \frac{16}{16+x}$

$=\frac{32}{5(16+x)}$

Getting two blue balls from first and second urn is $=\frac{6}{10}\times \frac{x}{16+x}$

$=\frac{3x}{5(16+x)}$

The probability that both balls are the same in color is $=\frac{32}{5(16+x)}+\frac{3x}{5(16+x)}$

Given that the probability that both balls are the same in color is 0.44.

According to the problem,

$\frac{32}{5(16+x)}+\frac{3x}{5(16+x)}=0.44$

$\Rightarrow \frac{32+3x}{5(16+x)} =0.44$

$\Rightarrow \frac {32+3x}{(80+5x)} =0.44$

$\Rightarrow 32+3x =0.44(80+5x)$

$\Rightarrow 32+3x =35.2 +2.2x$

$\Rightarrow 3x -2.2 x= 35.2 -32$

$\Rightarrow 0.8x= 3.2$

$\Rightarrow x = 4$

Therefore the of blue in the second urn is 4.

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2 years ago

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sukhopar [10]

Answer:

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Step-by-step explanation:

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2 years ago

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