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3 years ago

5. My red or blue shirt club has 32 members. If the number of people wearing read shirts

Mathematics

1 answer:

IgorC [24]3 years ago

7 0

Answer:

(red shirt people) r - 4 $\sqrt[3]{b}$ (blue shirt people)

Step-by-step explanation:

I don't know if you want an equation or an exact answer. I have an equation though.

I got this answer like this:

4 less is minus 4

triple is multiplication and/or times

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stepladder [879]

Set R(x) equal to C(x). Solve the resulting equation for x, paying attention to the domain of C(x):

Solve 200x-2x^2 = -x^2 + 25x + 31000 for x greater than or equal to zero but less than or equal to x. That means you must reject any solution outside of this domain.

Rearrange all of these terms on the left side of your equation and place zero on the right side. Next, rearrange all of the terms on the left side in descending order by powers of x. Next, solve the resulting quadratic equation. Make certain to reject any x that does not fall within [0,100].

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4 years ago

Translate the phrase into an algebraic expression.<br> The product of 9 and y

Alenkinab [10]

Answer:9*y or 9y

Step-by-step explanation:

7 0

3 years ago

How many solutions are there to the equation below?

Reil [10]

I think it’s B I’m not to sure but I know someone with that exact solution

5 0

3 years ago

PLEASE HELP!!

Bezzdna [24]

Answer:

$y=-\dfrac{9}{40}x^2+\dfrac{441}{40}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}$

Step-by-step explanation:

If a parabola has its vertex on the y-axis, then its equation is

$y=ax^2+b$

This parabola passes through the point R(3,9), then

$9=a\cdot 3^2+b\\ \\9=9a+b$

The area of the right triangle PQR is

$A_{PQR}=\dfrac{1}{2}\cdot PQ\cdot QR$

Find PQ and QR, if $P(x_1,0),\ Q(3,0),\ R(3,9):$

$PQ=\sqrt{(x_1-3)^2+(0-0)^2}=|x_1-3|\\\\QR=\sqrt{(3-3)^2+(0-9)^2}=9$

Now,

$40.5=\dfrac{1}{2}\cdot |x_1-3|\cdot 9\\ \\90=9|x_1-3|\\ \\|x_1-3|=10\\ \\x_1-3=10\ \text{or}\ x_1-3=-10\\ \\x_1=13\ \text{or }x_1=-7$

We get two possible points $P_1(-7,0)$ and $P_2(13,0).$

For point $P_1:\\$

$0=a\cdot (-7)^2+b\\ \\49a+b=0$

So,

$b=-49a\\ \\9=9a-49a\\ \\-40a=9\\ \\a=-\dfrac{9}{40}\\ \\b=\dfrac{441}{40}\\ \\y=-\dfrac{9}{40}x^2+\dfrac{441}{40}$

For point $P_2:\\$

$0=a\cdot (13)^2+b\\ \\169a+b=0$

So,

$b=-169a\\ \\9=9a-169a\\ \\-160a=9\\ \\a=-\dfrac{9}{160}\\ \\b=\dfrac{1,521}{160}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}$

5 0

3 years ago

PLEASE HELP!

Sveta_85 [38]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{2}{ h},\stackrel{-5}{ k})\qquad \qquad 
radius=\stackrel{12}{ r}
\\\\\\\
[x-2]^2+[y-(-5)]^2=12^2\implies (x-2)^2+(y+5)^2=144$

4 0

3 years ago