HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP

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:

huh i am very confused

Step-by-step explanation:

Answer: 9.13333333333

3 1/3+ 5 4/5=

Answer#2  1.86666666667

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: 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:

a = 9

Step-by-step explanation:

For solve the problem we can use the Pythagorean theorem in this way

a^2 + (3√3)^2 = (6√3)^2

If we solve the equation we have

a^2 + 27 = 108

a^2 = 81

a = +/-9

we take only the the positive value because a length can’t be negative

a = 9

Answer:

[tex]2x+3y\geq 60\\x>10[/tex]

Step-by-step explanation:

We know that she will be selling bracelets for $2 each and earrings for $3 each. That means we will have 2x+3y, where x is the number of bracelets she sells and y is the number of earrings she sells. Then we will have [tex]\geq 60[/tex] because she wants to sell at least $60 worth of earrings and bracelets. That means the amount she makes from bracelets and earrings combined has to equal 60 or more, thus we use the greater than or equal to sign.

Now we have [tex]x>10[/tex], where x is still the number of bracelets sold. She said she will sell more than 10, which means we will just have [tex]>10[/tex]. If she said she wanted to sell at least 10, then it would be [tex]\geq10[/tex].

Therefore the system of inequalities is:

[tex]2x+3y\geq60\\x>10[/tex]

Hope this helps!

Answer:

$15

Step-by-step explanation:

370-40=330

12+10=22

330÷22=15

15*22=330

Answer:

ok ok, i think it's the last one

Step-by-step explanation:

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]