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Dmitrij [34]
3 years ago
15

In the diagram, how many angles are alternate exterior angles with angle 5?

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
4 0

Answer:

Angle 11 is the alternate exterior of angle 5 meaning there is only 1 angle that is alternate of angle 5.

Step-by-step explanation:

An exterior angle is an angle outside of the shape (hence ex meaning out). However, an ALTERNATE angle is an angle opposite of a transversal line of another angle. The angles on the same transversal line of 5 are 8, 7, 9, 10, 12, and 11. However, we are looking for ALTERNATE EXTERIOR angles not just exterior. 8, 7, 9, 10, and 12 are just exterior which leaves angle 11 as alternate exterior. Angle 11 is the alternate exterior of angle 5 meaning there is only 1 angle that is alternate of angle 5.

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Answer:SORRY:(

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Which of the following best describes the graph of the polynomial function below?
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Answer:

A. The graph has two zeroes.

Step-by-step explanation:

A <em>zero of a polynomial</em> is a value of x that makes the polynomial equal to zero.

In other words, it is <em>an x-intercept</em>.

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Use the rational zero theorem to determine all possible rational zeros of f(x)=3x^3-2x^2+5x+9
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Step-by-step explanation:

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3 years ago
A ball is thrown up in the air, and it's height (in feet) as a function of time (in seconds) can be written as h(t) = -16t2 + 32
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Answer:

1) Using properties of the quadratic equation:

Here the height equation is:

h(t) = -16t^2 + 32t + 6

We can see that the leading coefficient is negative, this means that the arms of the graph will open downwards.

Then, the vertex of the quadratic equation will be the maximum.

Remember that for a general quadratic equation:

a*x^2 + b*x + c = y

the x-value of the vertex is:

x = -b/2a

Then in our case, the vertex is at:

t = -32/(2*-16) = 1

The maximum height will be the height equation evaluated in this time:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

2) Second method, using physics.

We know that an object reaches its maximum height when the velocity is equal to zero (the velocity equal to zero means that, at this point, the object stops going upwards).

If the height equation is:

h(t) = -16*t^2 + 32*t + 6

the velocity equation is the first derivation of h(t)

Remember that for a function f(x) = a*x^n

we have that:

df(x)/dx = n*a*x^(n-1)

Then:

v(t) = dh(t)/dt = 2*(-16)*t + 32 + 0

v(t) = -32*t + 32

Now we need to find the value of t such that the velocity is equal to zero:

v(t) = 0 = -32*t + 32

       32*t = 32

            t = 32/32 = 1

So the maximum height is at t = 1

(same as before)

Now we just need to evaluate the height equation in t = 1:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

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