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tekilochka [14]
3 years ago
11

Someone pls answer this it’s so confusing lol

Mathematics
1 answer:
GenaCL600 [577]3 years ago
4 0

Answer:

<h2>y = \frac{w}{h -  {2c}^{3} }</h2>

Step-by-step explanation:

w = yh - 2yc³

First of all factorize y out of the expression on the right side of the equation

That's

w = y( h - 2c³)

Next divide both sides by (h - 2c³) to make y stand alone

We have

<h3>\frac{y(h -  {2c}^{3}) }{h -  {2c}^{3} }  =  \frac{w}{h -  {2c}^{3} }</h3>

We have the final answer as

<h3>y = \frac{w}{h -  {2c}^{3} }</h3>

Hope this helps you

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Find the values of k so that each remainder is three. <br> 10. (x^2+ 5x + 7) = (x + k)
Goryan [66]

Answer:

k=1\text{ or } k=4

Step-by-step explanation:

We can use the Polynomial Remainder Theorem. It states that if we divide a polynomial P(x) by a <em>binomial</em> in the form (x - a), then our remainder will be P(a).

We are dividing:

(x^2+5x+7)\div(x+k)

So, a polynomial by a binomial factor.

Our factor is (x + k) or (x - (-k)). Using the form (x - a), our a = -k.

We want our remainder to be 3. So, P(a)=P(-k)=3.

Therefore:

(-k)^2+5(-k)+7=3

Simplify:

k^2-5k+7=3

Solve for <em>k</em>. Subtract 3 from both sides:

k^2-5k+4=0

Factor:

(k-1)(k-4)=0

Zero Product Property:

k-1=0\text{ or } k-4=0

Solve:

k=1\text{ or } k=4

So, either of the two expressions:

(x^2+5x+7)\div(x+1)\text{ or } (x^2+5x+7)\div(x+4)

Will yield 3 as the remainder.

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Rina8888 [55]

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If f(x) = StartFraction x Over 2 EndFraction + 8, what is f(x) when x = 10?
san4es73 [151]

Answer:

13

Step-by-step explanation:

You are given the function

f(x)=\dfrac{x}{2}+8

To find the value of the function f(x) at point x=a, you have to substitute a instead of x into the function expression.

In your case, you have to evaluate f(x) when x=10. Substitute 10 instead of x into the function expression:

f(10)=\dfrac{10}{2}+8=5+8=13

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3 years ago
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rosijanka [135]

Answer:

D 714

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210 times 3.40

is 714

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