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serg [7]
2 years ago
5

The graph shows y as a function of x

Mathematics
1 answer:
nirvana33 [79]2 years ago
3 0
I'm guessing Q lol. I hope that's helpful.
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PLEASE HELP 60 POINTSSSS
anastassius [24]

Answer:

x = 6.5 units

Step-by-step explanation:

The 2 given squares are the on the legs of the right triangle with hypotenuse x

Using Pythagoras' identity in the right triangle

The square on the hypotenuse is equal to the sum of the squares on the other 2 legs , then

x² = 13 + 29.25 = 42.25 ( take square root of both sides )

x = \sqrt{42.25} = 6.5 units

5 0
2 years ago
Find the value of c that makes t^2– 22t+ c a perfect square trinomial.
riadik2000 [5.3K]

Answer:

ZkgJHhHJ HHha sorry i am not sure

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B9x%2B20%3D0" id="TexFormula1" title="x^{2} +9x+20=0" alt="x^{2} +9x+20=0" ali
Hoochie [10]

Answer:

We have been given an equation x^2+9x+20=0

We will factorize it by  middle term splitting.

Step 1: x^2+5x+4x+20=0

Taking common factors out by clubing first two terms ad last two terms

Step 2: x(x+5)+4(x+5)=0

Step 3: (x+4)(x+5)=0

Equating both above factors to zero as:

Step 4: (x+4)=0\Rightarrow x=-4

   And (x+5)=0\Rightarrow x=-5.

3 0
3 years ago
Read 2 more answers
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
A city that has an elevation of 15 meters is closer to sea level than a city that has an elevation of -10 meters.
Darya [45]

Answer:

Step-by-step explanation: Your answer will be -10 because even though -10 is much lower than 15 it’s still closer to sea level.

3 0
2 years ago
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