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Yakvenalex [24]
3 years ago
9

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

Mathematics
1 answer:
STatiana [176]3 years ago
7 0

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!

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Suppose lines EF and GH are reflected over the line Y equals X to form the lines JK and LM which statement would be true about t
katrin [286]

Answer:

A) JK and LM will be parallel to each other.

Step-by-step explanation:

On reflection on y=x line the x co-ordinate changes with y co-ordinate and y co-ordinate changes with x co-ordinate

(x,y)\rightarrow (y,x)

Points on line EF

(0,6) , (-5,-2)

On reflection of this line on y=x the new points we get for line JK are

(6,0),(-2,-5)

Points on line GH

(-4,9),(-9,1)

On reflection on y=x line the new points we get for line LM are

(9,-4),(1,-9)

Slope of line JK

m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-5)-0}{(-2)-6} \\m=\frac{-5}{-8}=\frac{5}{8}

Slope of line LM

m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-9)-(-4)}{1-9} \\m=\frac{-9+4}{-8}=\frac{-5}{-8}\\m=\frac{5}{8}

For two line to be parallel, their slopes will be same.

m_{JK} =\frac{5}{8} , m_{LM}=\frac{5}{8}

Since slopes of lines JK and LM are same therefore we can say that these are parallel to each other.

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Answer:

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What is the sum of (8a+2b-4) and (3b-5)
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Answer:

8a+5b-9

Step-by-step explanation:

(8a+2b-4)+(3b-5)

8a+2b+3b-4-5

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