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2 years ago

The altitude of the base of a cloud

Mathematics

1 answer:

natima [27]2 years ago

7 0

The height of the ceiling is 162.31 meters

The situation will form a right angle triangle.

<h3>Right angle triangle:</h3>

Right triangle has one of its angles as 90 degree. Therefore, the sides can be found using trigonometric ratios

The triangle formed, the theodolite distance from the spotlight is the adjacent side. Therefore,

tan 62.7 = opposite / adjacent

tan 62.7 = x / 83

cross multiply

x = 83 × tan 62.7

x = 160.809554025 m

Therefore, the height of the ceiling form the ground is as follows:

160.809554025 + 1.5 = 162.309554025 m

learn more on right angle triangle here; brainly.com/question/13481976

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SO first off you wanna make both numbers into mix factions. 6 8/9 will be 62/9 and 2 1/2 will be 5/2. Next you wanna multiply straight across. so now the answe should be 310/18 then simply for a mixed number. you can figure that out

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The answer to the question is -16.

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Jenna grabs a sample of trail mix from a bag. The mix contains peanuts, almonds, raisins, cashews, and chocolate chips. She want

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10 5 40

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2 years ago

What is the length of x?

Marat540 [252]

Answer:

The length of 'x' is 4

Step-by-step explanation:

We can use the pythagorean theorem to solve this.

The pythagorean theorem is:

$a ^{2} + b^{2} = {c}^{2}$

$a ^{2} + b^{2} = c^{2} \\ {3}^{2} + {b}^{2} = 5^{2} \\ 9 + {b}^{2} = 25 \\ \frac{ - 9 \: \: \: \: \: \: \: \: = - 9}{b ^{2} = \sqrt{16} } \\ b = 4$

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2 years ago

(3x-5)/(x-5)>=0 <br> How do I get the answers for this problem?

Diano4ka-milaya [45]

$\frac{3x-5}{x-5}$ > 0

First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like: $\frac{3x-5}{x-5}$ = 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x = $\frac{5}{3}$
Sixth, from the values of x above, we have these 3 intervals to test:
x < $\frac{5}{3}$
$\frac{5}{3}$ < x < 5
x > 5
Seventh, pick a test point for each interval.

1. For the interval x < $\frac{5}{3}$ :
Let's pick x - 0. Then, $\frac{3x0-5}{0-5}$ > 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.

2. For the interval $\frac{5}{3}$ < x < 5:
Let's pick x = 2. Then, $\frac{3x2-5}{2-5}$ > 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.

3. For the interval x > 5:
Let's pick x = 6. Then, $\frac{3x6-5}{6-5}$ > 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x < $\frac{5}{3}$ and x > 5

Answer: x < $\frac{5}{3}$ and x > 5

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3 years ago