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wlad13 [49]
2 years ago
8

The hypotenuse of a 90-45-45 triangle is 20√2. What is the short leg and the long leg>

Mathematics
1 answer:
stepan [7]2 years ago
6 0

Answer:

Both legs would be 20

Step-by-step explanation:
In the image below, it shows the relationship between the legs and the hypotenuse in every 45 45 90 triangle. If we know that the hypotenuse is 20, then the 2 legs are both 20.

I am slightly confused with the problem since they asked for the "short" leg and the "long" leg. Both legs should be the same length because a 45 45 90 triangle is isosceles, with the two legs being the same length.

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Simplify.<br><br> 100‾‾‾‾√<br><br> 20<br><br> 10 <br><br> −1<br><br> −25
olga_2 [115]

Answer:

10

Step-by-step explanation:

\sqrt{100} =\sqrt{(10)(10)}  = 10

 

8 0
2 years ago
Solve for y.<br> 8+4y= -8
Andrew [12]

Answer:

-4

Step-by-step explanation:

8+4y= -8

4y= -8 -8

4y= -16

divide both sides by the coeffient of the variable. in this case divide both sides by 4

y=-4

7 0
3 years ago
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4 Write an inequality for the word phrase<br> -5 is fewer than 6 less than a numberk
olchik [2.2K]

Answer:

6 - 5 < X

Is your expression

5 0
3 years ago
Which of the following is equal to the expression below? (160x243)^1/5
Kipish [7]
160 = 2 x 2 x 2 x 2 x 2 x 5 = 2^5 x 5
243 = 3 x 3 x 3 x 3 x 3 = 3^5
So
(160 * 243)^1/5
= 5th root of (160 * 243)
= 5th root of (2^5 * 5 * 3^5)
= 2 * 3 * (5th root of 5)
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3 0
3 years ago
Sarah kicked a ball in the air. The function
Aleksandr-060686 [28]

Answer: The ball hits the ground at 5 s

Step-by-step explanation:

The question seems incomplete and there is not enough data. However, we can work with the following function to understand this problem:

f=30 t- 6t^{2} (1)

Where f models the height of the ball in meters and t the time.

Now, let's find the time t when the ball Sara kicked hits the ground (this is when f=0 m):

0=30 t- 6t^{2} (2)

Rearranging the equation:

6t^{2}-30 t=0 (3)

Dividing both sides of the equation by 6:

t^{2}-5 t=0 (4)

This quadratic equation can be written in the form at^{2}+bt+c=0, and can be solved with the following formula:  

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} (5)  

Where:  

a=1  

b=-5  

c=0  

Substituting the known values:  

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(0)}}{2(1)} (6)  

Solving we have the following result:

t=5 s  This means the ball hit the ground 5 seconds after it was kicked by Sara.

6 0
3 years ago
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