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2 years ago

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At SMS, students were encouraged to wear school colors of blue and white. 900 Students wore their school colors while 300 studen

ts did not. What percent of the school population did not wear school colors?

1 answer:

emmasim [6.3K]2 years ago

4 0

Answer:300/1200

Step-by-step explanation:part/whole = %/100

300/1200 =25

/100

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Find an equation, or a set of equations, to describe the set of points that are equidistant from the points p(−9, 0, 0) and q(3,

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X = -3.
The distance from p(-9, 0, 0) is
d = sqrt((x+9)^2 + y^2 + z^2)
The distance from q(3,0,0) is
d = sqrt((x-3)^2 + y^2 + z^2)
Let's set them equal to each other.
sqrt((x+9)^2 + y^2 + z^2) = sqrt((x-3)^2 + y^2 + z^2)
Square both sides, then simplify
(x+9)^2 + y^2 + z^2 = (x-3)^2 + y^2 + z^2
x^2 + 18x + 81 + y^2 + z^2 = x^2 - 6x + 9 + y^2 + z^2
18x + 81 = - 6x + 9
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x = -3
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X^2+4x-1/(x+2)^2=0 solve for x

dalvyx [7]

<span>x^2+4x-1/(x+2)^2=0
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$\frac{x^2+10x-7}{x^2+10x+25} =0\\x^2+10x-7=0\\x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ;
where a = 1, b = 4 and c = -1

$x=\frac{-4\pm\sqrt{4^2-(4\times1\times(-1)}}{2\times1}\\=\frac{-4\pm\sqrt{16+4}}{2}=\frac{-4\pm\sqrt{20}}{2}\\=\frac{-4\pm\sqrt{4\times5}}{2}=\frac{-4\pm2\sqrt{5}}{2}=-2\pm\sqrt{5}\\=-2+\sqrt{5}
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Each week, Heather’s company has $5000 in fixed costs plus an additional $250 for each system produced. The company is able to p

kvv77 [185]

The question is an illustration of composite functions.

Functions c(n) and h(n) are $\mathbf{c(n) = 5000 + 250n}$ and $\mathbf{n(h) = 5h}$
The composite function c(n(h)) is $\mathbf{c(n(h)) = 5000 + 1250h}$
The value of c(n(100)) is $\mathbf{c(n(100)) = 130000}$
The interpretation is: <em>"the cost of working for 100 hours is $130000"</em>

The given parameters are:

$5000 in fixed costs plus an additional $250
5 systems in one hour of production

<u>(a) Functions c(n) and n(h)</u>

Let the number of system be n, and h be the number of hours

So, the cost function (c(n)) is:

$\mathbf{c(n) = Fixed + Additional \times n}$

This gives

$\mathbf{c(n) = 5000 + 250 \times n}$

$\mathbf{c(n) = 5000 + 250n}$

The function for number of systems is:

$\mathbf{n(h) = 5 \times h}$

$\mathbf{n(h) = 5h}$

<u>(b) Function c(n(h))</u>

In (a), we have:

$\mathbf{c(n) = 5000 + 250n}$

$\mathbf{n(h) = 5h}$

Substitute n(h) for n in $\mathbf{c(n) = 5000 + 250n}$

$\mathbf{c(n(h)) = 5000 + 250n(h)}$

Substitute $\mathbf{n(h) = 5h}$

$\mathbf{c(n(h)) = 5000 + 250 \times 5h}$

$\mathbf{c(n(h)) = 5000 + 1250h}$

<u>(c) Find c(n(100))</u>

c(n(100)) means that h = 100.

So, we have:

$\mathbf{c(n(100)) = 5000 + 1250 \times 100}$

$\mathbf{c(n(100)) = 5000 + 125000}$

$\mathbf{c(n(100)) = 130000}$

<u>(d) Interpret (c)</u>

In (c), we have: $\mathbf{c(n(100)) = 130000}$

It means that:

The cost of working for 100 hours is $130000

Read more about composite functions at:

brainly.com/question/10830110

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