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hichkok12 [17]
2 years ago
13

Need help with Trigonometry will mark brainest!!

Mathematics
1 answer:
Alexxx [7]2 years ago
5 0

The height of the kite above the ground will be 108.22m.

<h3>How to get the height?</h3>

From the information, the wind makes an angle of 56° to the ground and the kite is about 73m.

Therefore, the height of the kite above the ground will be:

Tan 56° = opposite/adjacent

1.48256 = x/73

x = 73 × 1.48256

x = 108.22m

Learn more about angles on:

brainly.com/question/25716982

#SPJ1

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What is the correlation coefficient for the data shown in the table?
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Answer:

The correlation coefficient is -1

Step-by-step explanation:

Here, we want to get the correlation coefficient for the given dataset

from the table, what do we observe?

As x is increasing by 5 units;

y is decreasing by 5 units

What this mean is that there is a negative correlation between both

Thus, what we have here is that, since the rate of increase/decrease on both ends is of equal magnitude but opposite signs, we can conclude that the correlation coefficient is -1

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Which of the following are true statements?
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Answer:

I think the answer is B I guessed it.

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A 20 foot ladder leaning against a wall is used to reach a window that is 17 feet above the ground. How far from the wall is the
Liono4ka [1.6K]

Answer:

The wall is 10.5 foot far from the bottom of the ladder.

Step-by-step explanation:

Given:

Height of the ladder leaning against wall (Hypotenuse) = 20 foot.

Height from where the window is above from the ground (Leg1) = 17 feet.

To find the distance of the bottom of the ladder far from the wall (Leg2) = ?

Now, by using the pythagorean theorem:

Hypotenuse^{2} = (Leg1)^{2} + (Leg2)^{2}

20^{2} =17^{2}+(Leg2) ^{2}

400=289+(Leg2)^{2}

400-289=(Leg2)^{2}

111=(Leg2)^{2}

by squaring both sides

Leg2=10.535

10.535 foot rounded nearest tenth is 10.5 foot.

Therefore, the wall is 10.5 foot far from the bottom of the ladder.  

6 0
3 years ago
HELP ME PLEASEEEEEEEEEE
Licemer1 [7]

Answer:

The area is 35

The perimeter is 24

Step-by-step explanation:

You can count the individual units of the length and the width of the rectangle.

The width is 5, the length is 7

The area is 5*7=35

The perimeter is 5+5+7+7=24

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3 years ago
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<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{
Jlenok [28]

Answer:

x>2

Step-by-step explanation:

When given the following inequality;

(x^2+x-3):(x^2-4)\geq1

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

\frac{x^2+x-3}{x^2-4}\geq1

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

\frac{x^2+x-3}{x^2-4}-1\geq0

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0

Perform the operation on the other side distribute the negative sign and combine like terms;

\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0

Factor the equation so that one can find the intervales where the inequality is true;

\frac{x+1}{(x-2)(x+2)}\geq0

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

-1, 2, -2

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

x\leq-2\\-2

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

x\leq-2   -> negative

-2   -> neagtive

-1\leq x   -> neagtive

x>2   -> positive

Therefore, the solution to the inequality is the following;

x>2

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