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goldenfox [79]
3 years ago
5

Which of the following equations is equivalent to x - y = 8?

Mathematics
2 answers:
matrenka [14]3 years ago
4 0
Say x = 12 and y = 4 
12 - 4 = 8

Number 1
2(12) - 4(4) = 8
24 - 16 = 8
Number 1 is your Answer
timofeeve [1]3 years ago
3 0

Answer:

The equivalent equation is 1) 2x - 4y = 8

Step-by-step explanation:

The given equation is x - y = 8.

Here we have 8 on the right hand side.

Now find the equation which has 8 on the right hand side.

It is 2x - 4y = 8

Therefore, the equivalent equation is 1) 2x - 4y = 8

Now let's find the value of x and y.

From the equation x - y = 8, we get

x = y + 8

Now plug in x = y + 8 in the equation 2x - 4y = 8

2(y + 8) -4y = 8

Use the distributive property

2y + 16 -4y = 8

-2y + 16 = 8

-2y = 8 -16

-2y  = -8

Dividing both sides by -2, we get

y = 4

Now plug in y = 4 in x = y + 8 to get the value of x.

x = 4 + 8 = 12

So the value of x = 12 and y = 4

If we plug these values in the above two equations, we get will get 8.

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Solve for b.<br> 7<br> b<br> 12<br> b= ✓ [?]<br> Pythagorean Theorem: a2 + b2 = c2<br> Enter
fomenos

Answer: b=\sqrt{95}

Step-by-step explanation:

To solve for b, we want to use the Pythagorean Theorem as given.

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A lacrosse player throws a ball into the air from a height of 6 feet with an initial vertical velocity of 64 feet per second. Wh
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Answer:

Step-by-step explanation:

I'm going to use calculus to solve this, because it's the simplest way.  

The acceleration due to gravity in feet is the second derivative of the position function.  We will start with the acceleration and work backwards with antiderivatives to get to the position function.

a(t) = -32.  Going backwards and using the fact that the initial vertical velocity is 64 ft/sec, our velocity function is

v(t) = -32t + 64.  Going backwards and using the fact that the initial height of the ball is 6 feet, our position function is

s(t)=-16t^2+64t+6

The first part of this question asks us the maximum height of the ball.  From Physics, we learn that the maximum height of a projectile is reached when the velocity is 0, which happens to be right where the projectile stops for a nanosecond in the air to turn around and come back down.  We set the velocity function equal to 0 and solve for t.

0 = -32t + 64 and

0 = -32(t - 2).  By the Zero Product Property, either -32 = 0 or t - 2 = 0.  It's obvious that -32 does not equal 0, so t - 2 must equal 0.  Solving this for t:

t - 2 = 0 so

t = 2 seconds.  Since the maximum height is reached at a time of 2 seconds, we plug 2 seconds into the position function to get its position at 2 seconds (which is also the max height of the ball).

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s(2) = 70 feet

Now we want to know when the ball will hit the ground.  "When" is a time value, and we know that the height of the ball on the ground is 0, so we sub in a 0 for s(t) and factor the quadratic.

Using the quadratic formula:

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t=\frac{-64+\sqrt{4480} }{-32} and

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Again, from Physics, we know that a projectile reaches it max height at halfway through its travels, so it just goes to follow logically that if it halfway through its travels at 2 seconds, then it will hit the ground at 4 seconds.  And it does!! How awesome is that?!

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