Which number line shows the correct locations of all the given values?

Answer:

I believe one flash drive would cost $16.50.

Step-by-step explanation:

55=2x+22

-22       -22

33=2x

/2    /2

16.50=x

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Answer:

$15

Step-by-step explanation:

370-40=330

12+10=22

330÷22=15

15*22=330

Answer:

a = 1.4 and d = 6.9

Step-by-step explanation:

To get the value of d and a, we will use the SOH CAH TOA identity

For the first tringle;

Opposite = d ((opposite angle 60 degrees)

Adjacent = 4

Tan theta = opp/adj

Tan 60 = d/4

d  = 4tan60

d = 4(1.732)

d = 6.9

For the second triangle;

Opposite = a

Hypotenuse = 2

Since sin theta = opp/hyp

Sin45 = a/2

a = 2sin45

a = 2(0.7071)

a = 1.4

Hence a = 1.4 and d = 6.9

Answer:

B.12

Step-by-step explanation:

For your first draw you have 4 possible options, then on your second draw you have 3 possible options since you took out a pair. So to figure out the number of possible outcomes you simply multiply the possibilities by each other to get the number of possible outcomes.  

Answer:

Step-by-step explanation:

Nothing further can be done with this topic.  

f ( x ) = 3 x 2 − 10 x − 9

Answer:

The maximum height of the ball is 380.25 feet in the air.

Step-by-step explanation:

The quadratic function:

[tex]h(t)=-16t^2+132t+108[/tex]

Models the ball's height h(t), in feet, above the ground t seconds after it was thrown.

We want to determine the maximum height of the ball.

Note that this is a quadratic function. Therefore, the maximum or minimum value will always occur at its vertex point.

Since our leading coefficient is leading, we have a maximum point. So to find the maximum height, we will find the vertex. The vertex of a quadratic equation is given by:

[tex]\displaystyle \left(-\frac{b}{2a},f\left(\frac{b}{2a}\right)\right)[/tex]

In this case, a = -16, b = 132, and c = 108. Find the t-coordinate of the vertex:

[tex]\displaystyle t=-\frac{132}{2(-16)}=-\frac{132}{-32}=\frac{33}{8}=4.125[/tex]

So, the maximum height occurs after 4.125 seconds of the ball being thrown.

To find the maximum height, substitute this value back into the equation. Thus:

[tex]h(4.125)=-16(4.125)^2+132(4.125)+108=380.25\text{ feet}[/tex]

The maximum height of the ball is 380.25 feet in the air.

Answer:

-5FStep-by-step explanation:

10-5x3=-5 F