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olga55 [171]
2 years ago
12

In ΔHIJ, the measure of ∠J=90°, the measure of ∠H=71°, and JH = 2 feet. Find the length of IJ to the nearest tenth of a foot.

Mathematics
1 answer:
Dmitrij [34]2 years ago
3 0

Answer:

IJ is 5.8 ft

Step-by-step explanation:

Here, we want to calculate the length of IJ

to Find IJ, we need the appropriate trigonometric ratio

The trigonometric ratio to use in this case is the one that connects the opposite and the adjacent

from the question, IJ is the opposite

This is the side facing the given angle

HI is the hypotenuse which is the side facing the right-angle

HJ is the third side which is the adjacent

Mathematically;

That is the Tan

Tan 71 = IJ/2

IJ = 2 Tan 71

IJ = 5.8

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What is the direct variation equation if y varies directly with x and y = –6 when x = –36?
Vinil7 [7]
Direct variation is
y=kx
subsitute
-6=-36k
diivde both sdies by -36
1/6=k

the equation is
y=1/6x or B
8 0
3 years ago
14 + 5x = 3(-x + 3) -11 please show work step by step
mr Goodwill [35]

14+5x=3(-x+3)-11

14+5x=-3x+9-11

14+5x=-3x-2

add 2 to both sides

16+5x=-3x

subtract 5x from both sides

16=-8x

divide both sides by -8

-2=x

Hope this helps! :)

3 0
3 years ago
we need any help on this home I can you turn it in by Monday so can you help me please. i need on 9 and 10
marta [7]
9) yes, he is correct.   

50% as a decimal would be 0.50
7% as a decimal would be 0.07

10)Buy and Bye has a greater discount because 30% as a decimal is 0.30 and 1/4 as a decimal is 0.25.

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7 0
3 years ago
What is the range of the equation
a_sh-v [17]

The range of the equation is y>2

Explanation:

The given equation is y=2(4)^{x+3}+2

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

y=2 \cdot 4^{x+3}+2

This can be written as y=2^{1+2(x+3)}+2

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

x=2^{1+2(y+3)}+2

Solving for y, we get;

x-2=2^{1+2(y+3)}

Applying the log rule, if f(x) = g(x) then \ln (f(x))=\ln (g(x)), then, we get;

\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)

Simplifying, we get;

(1+2(y+3)) \ln (2)=\ln (x-2)

Dividing both sides by \ln (2), we have;

2 y+7=\frac{\ln (x-2)}{\ln (2)}

Subtracting 7 from both sides of the equation, we have;

2 y=\frac{\ln (x-2)}{\ln (2)}-7

Dividing both sides by 2, we get;

y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}

Let us find the positive values for logs.

Thus, we have,;

x-2>0

     x>2

The function domain is x>2

By combining the intervals, the range becomes y>2

Hence, the range of the equation is y>2

7 0
3 years ago
Help on questions 37 and 38 please!
jeka94

Answer:

Step-by-step explanation:

38.

M=\frac{1}{2} (x_{1}+x_{2})\\2M=x_{1}+x_{2}\\x_{1}=2M-x_{2}

37.

S=\frac{ab^2}{3} \\3S=ab^2\\b^2=\frac{3S}{a} \\b=\pm\sqrt{\frac{3S}{a} }

6 0
3 years ago
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