You wash 12 cars on Friday and 10 cars on Saturday for our fund-raiser you were also given a $40 donation from somebody who didn't want a car wash if you raised a total of $370 how much money did you earn per car wash

Answer:

ok ok, i think it's the last one

Step-by-step explanation:

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:

r=2.5mm

Step-by-step explanation:

Using the formula

[tex]\huge\underline{ \sf \: Formulae \: to \: remember \: for \: solving \: the \: question:}[/tex]

[tex] \\{ \sf radius \: of \: a \: circle = \frac{d}{2} where \: d \: is \: the \: diameter \: of \: the \: circle}[/tex]

[tex]\huge\underline{\sf Given:}[/tex]

[tex] \ \: { \sf The \: diameter \: of \: a \: circle \: is \: 5 mm.}[/tex]

[tex]\huge\underline{\sf To \: find \: out:}[/tex]

[tex] \ { \sf Radius \: of \: the \: circle}[/tex]

[tex]\huge\underline{\sf Solution:}[/tex]

[tex] \\{ \sf The \: diameter \: of \: the \: circle \: is \: 5 \: mm.}[/tex]

[tex] \\{ \sf Therefore \: radius \: of \: the \: circle}[/tex]

[tex]\\{ \sf = \frac{5}{2} \: mm} [/tex]

[tex] \\{ \sf = 2.5 \: mm}[/tex]

[tex]\huge\underline{\sf Answer:}[/tex]

[tex]\\{\sf The \: radius \: of \: the \: circle \: is \: 2.5 \: mm.}[/tex]

Answer: 61.58

Step-by-step explanation:

C = 2 π r = 2 · π · 9.8

π = pi

radius times pi (3.14) times 2 (because two radius makes the diameter)

9.8 times 3.14 times 2 = 61.57522in

Now we round.. 61.58

:) I hope this helps!  

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.