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3 years ago

12

Hello can anyone help me with my algebra i'm behind

1 answer:

madreJ [45]3 years ago

3 0

Answer:

sure

Step-by-step explanation:

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I think the answer's C: $7000/$8000
That's a surprisingly easy question for high school

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How does the multiplicity of a zero determine the behavior of the graph at that zero?

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2 years ago

Please help me!<br> Find the number x such that f(x) =1

STatiana [176]

Answer:

D

Step-by-step explanation:

We have the piecewise function:

$f(x) = \left\{ \begin{array}{ll} -\frac{1}{2}x-1 & \quad x \leq -2 \\ x & \quad x > -2 \end{array} \right.$

And we want to find x such that f(x)=1.

So, let's substitute 1 for f(x):

$1 = \left\{ \begin{array}{ll} -\frac{1}{2}x-1 & \quad x \leq -2 \\ x & \quad x > -2 \end{array} \right.$

This has two equations. So, we can separate them into two separate cases. Namely:

$1=-\frac{1}{2}x-1\text{ or } 1=x$

Let's solve for x in each case.

Case I:

We have:

$1=-\frac{1}{2}x-1$

Add 1 to both sides:

$2=-\frac{1}{2}x$

Let's cancel out the fraction by multiplying both sides by -2. So:

$2(-2)=(-2)\frac{-1}{2}x$

The right side cancels:

$-4=x\\$

Flip:

$x=-4$

So, x is -4.

Case II:

We have:

$1=x$

Flip:

$x=1$

This is the solution for our second case.

So, we have:

$x_1=-4\text{ or } x_2=1$

Now, can check to see if we have to to remove solution(s) that don't work.

Note that x=-4 is the solution to our first equation.

The first equation is defined only if x is less than -2.

-4 <em>is </em>less than -2. So, x=-4 is indeed a solution.

x=1 is the solution to our second equation.

The second equation is defined only if x is greater than or equal to -2.

1 <em>is</em> greater than or equal to -2. So, x=1 is <em>also </em>a solution.

Therefore, our two solutions are:

$x_1=-4\text{ or } x_2=1$

Out of our answer choices, we can pick D.

And we're done!

7 0

3 years ago

Need help plz i show me how to do it plz :)

DaniilM [7]

I’m pretty sure you multiply the sides your you find the formulas for the hight and width

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