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ozzi
3 years ago
7

PLEASE ANSWER ASAP!!!!

Mathematics
2 answers:
goldenfox [79]3 years ago
5 0

Answer:

she did nothing wrong lowkey

Step-by-step explanation:

kondor19780726 [428]3 years ago
5 0
The answer is -1 1/4 in decimal it would be -1.25
Hope it helps!
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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th
zimovet [89]

Answer:

883.3

Step-by-step explanation:

thats all.

8 0
3 years ago
Can someone please tell me the answer to these 2 problems. It has to doo with percentages
sweet-ann [11.9K]
1.) 100-82=18 18%=.18= 18/100= 9/50 2.) I need to see the table with the info to help
6 0
3 years ago
Maya wants to buy a new air conditioner. She could pay $149 in cash or make 4 payments of $39 each. How much would she save if s
Semenov [28]

Answer:

$7

Step-by-step explanation:

4 x 39 = 156

156 - 149 = 7

7 0
2 years ago
A volleyball court measures 30-feet wide by 60-feet long. The net is located 30 feet from the serve line. Assume the server stan
AveGali [126]

This is a tough question to answer, not because it's hard, but because it's hard to know what's already been taught.  For example we can express the components of the velocity off the racket in polar or rectangular form.  We could write the equation to include air resistance, definitely a concern for a real tennis serve, but not usually for high school math.

It's a projectile; there are really two variables, the initial velocity off the racket in the horizontal direction, and the initial velocity of the racket in the vertical direction.

Let's say the server hits the ball at her usual service velocity v₀.  Let's say the ball is at height h when it comes in contact with the racket. We'll further assume she gets to control the angle of elevation A she hits the ball.

In the horizontal direction we have constant velocity.

x = ( v₀ cos θ ) t

For us, we're interested in the time the ball takes to travel 31 horizontal feet.  We set x=31 feet.

t = x /  ( v₀ cos θ )  = 31 /  ( v₀ cos θ )

In the y direction we have motion in a constant gravitational field:

y = -16t² + ( v₀ sin θ ) t + h

-16 (feet per second per second) is the acceleration of gravity.  v₀ sin θ  is the vertical component of velocity.  h is the initial height of the ball when it hit the racket.

We need y > 7.5 to clear the net.  Let's set it equal to 7.5 to calculate the 'just barely' numbers.

7.5 = -16t² + ( v₀ sin θ ) t + h

We substitute  t = 31 /  ( v₀ cos θ )

7.5 - h = -16(31 /  ( v₀ cos θ ))² + ( v₀ sin θ ) ( 31 /  ( v₀ cos θ ) )

7.5 - h = -16(31²) /( v₀² cos²θ ) + 31 v₀ tan θ

Let's stop here; that's our equation, we don't need to solve it.  If we fixed h and v₀ this would be a trig problem to solve for θ, the angle she needs to hit the ball given her height and strength.

I've been documenting along the way; to recap, 7.5 is the height in feet of the net, h is the height of the ball when hit, -16 is the acceleration of gravity, 31 is the horizontal distance to the net, v₀ is the speed of the ball off the racket, θ<em> </em>is the angle of elevation of the ball off the racket.

3 0
3 years ago
How do i solve this?
vredina [299]
Multiply straight across
6 0
3 years ago
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