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Vika [28.1K]
2 years ago
9

Which of the following describes what happens when you change the a value in the equation shown below?

Mathematics
2 answers:
Over [174]2 years ago
7 0
I will say that it is C
Snezhnost [94]2 years ago
5 0

Answer:

Step-by-step explanation:

Changing the "a" value changes x-intercepts because when you distribute the "a" value to the binomials or the factored pairs to the right then the suitable x value which makes the quadratic equation also changes.

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sleet_krkn [62]

Answer:

Here for points

Step-by-step explanation:

3 0
2 years ago
What is the answer to this?
ehidna [41]

The answer is B, congruent

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3 years ago
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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
-Dominant- [34]

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).

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

s(2) = -64 + 128 + 6 so

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:

t=\frac{-64+/-\sqrt{4096-4(-16)(6)} }{-32} and

t=\frac{-64+/-\sqrt{4480} }{-32} which gives us the 2 solutions

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

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

Plugging into your calculator, the first t = -.0916500 and the second t = 4.091

We all know that time cannot ever be negative, so our t value is 4.09.

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?!

4 0
3 years ago
Given that cosΘ = -square root 2/2 and that Θ lies in quadrant III, determine the value of sinΘ.
Reptile [31]

Answer:

  C) (square root 2)/2

Step-by-step explanation:

In quadrant III, the sine of an angle is positive. The value will be ...

  sin(Θ) = √(1 -cos(x)^2) = √(1 -(-√2/2)^2) = √(2/4)

  sin(Θ) = (√2)/2

6 0
2 years ago
What is the answer to a and b please and thank you
vagabundo [1.1K]
A.) according to the graph, Austin appears to be burning 10 calories per minute. If you look at a perfect point on the graph, which I chose (50,5) and you do x over y or calories per minute, you get a unit rate of 10 calories per minute.

b.) Since you have already found the unit rate in question a, your slope would be 10. 
5 0
3 years ago
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