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Nikitich [7]
3 years ago
6

Given that JL is perpendicular KL and KH is congruent HL, find KL

Mathematics
2 answers:
Vlad [161]3 years ago
8 0

Answer:

KL = 7.8

Step-by-step explanation:

KH = HL

KH + HL = KL

3.9 + 3.9 = 7.8

avanturin [10]3 years ago
7 0

Answer:

KL=7.8

Step-by-step explanation:

We have been given a kite. We are asked to find the length of segment KL.

We are told that KH is congruent HL.

We can see that length of KL is equal to KH plus HL.

KL=KH+HL

KL=KH+KH

KL=2KH

KL=2(3.9)

KL=7.8

Therefore, the length of KL is 7.8 units.

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X-11/2x+10÷x^2-8×-33/x+5​
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Answer:

Step-by-step explanation:

SOLVE FOR X

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x=  

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≈−3.302775638

Hope this helps xxx

6 0
3 years ago
Read 2 more answers
What is x in 3x+3= -2x-12
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3x+3=-2x-12 \\ 3x+2x=-12-3 \\ 5x=-15 \\ x=-3 \\ \\ \text{The answer is x=-3 ;)}
7 0
3 years ago
Read 2 more answers
1. whats 2/3 plus 1/9<br> 2. 5/7 plus 1/2<br> 3.7/15 plus 2/15
Oksana_A [137]

1. 5/9
2. 17/14
3. 9/15
7 0
4 years ago
Read 2 more answers
Please help! Urgent!
LenKa [72]

Answer:

x = 127 degrees

Step-by-step explanation:

By the external angle of a triangle theorem:-

x + 5 = 79 + m < ACB

m < ACB = 180 - x, so:-

x + 5 = 79 + 180 - x

2x = 79 + 180 - 5 = 254

x = 127  ( answer)

3 0
3 years ago
Cos(theta) = - 3/4 and is in the 3rd quadrant, find the following:
elena-14-01-66 [18.8K]

Answer:

\begin{gathered} sin(\theta)=\frac{-\sqrt{7}}{4} \\ cos(\theta)=-\frac{3}{4} \\ tan(\theta)=\frac{\sqrt{7}}{3} \\ csc(\theta)=-\frac{4}{\sqrt{7}} \\ sec(\theta)=-\frac{4}{3} \\ cot(\theta)=\frac{3}{\sqrt{7}} \end{gathered}

Step-by-step explanation:

If theta is in the third quadrant, draw the diagram to easily identify the other trigonometric relations:

Solve for the missing leg of the triangle, using the Pythagorean theorem:

\begin{gathered} \text{ adjacent}^2+\text{ opposite}^2=\text{ hypotenuse}^2 \\ -3^2+\text{ opposite}^2=4^2 \\ \text{ opposite=}\sqrt{16-9} \\ \text{ opposite=}\sqrt{7} \end{gathered}

Therefore, for the trigonometric relationships:

\begin{gathered} \text{ sin\lparen}\theta)=\frac{opposite}{hypotenuse} \\ \text{ cos\lparen}\theta)=\frac{adjacent}{hypotenuse} \\ tan(\theta)=\frac{opposite}{adjacent} \\ csc(\theta)=\frac{hypotenuse}{opposite} \\ sec(\theta)=\frac{hypotenuse}{adjacent} \\ cot(\theta)=\frac{adjacent}{opposite} \end{gathered}

Now, substitute and solve for the relations:

\begin{gathered} sin(\theta)=\frac{-\sqrt{7}}{4} \\ cos(\theta)=-\frac{3}{4} \\ tan(\theta)=\frac{\sqrt{7}}{3} \\ csc(\theta)=-\frac{4}{\sqrt{7}} \\ sec(\theta)=-\frac{4}{3} \\ cot(\theta)=\frac{3}{\sqrt{7}} \end{gathered}

7 0
1 year ago
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