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svetoff [14.1K]
3 years ago
5

How can you get coefficients that are equal for X? 3x + 2y = 12 6x + 3y = 21

Mathematics
1 answer:
Verizon [17]3 years ago
5 0

Step-by-step explanation:

To make the coefficient of x equal we can multiply both sides of (1) by 2. the answer is B

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Step-by-step explanation:

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Question 2 Multiple Choice Worth 1 points)
Sati [7]

Comparing g(x) with f(x), you can see that the function f(x) is translated to the right by 6 units to produce g(x) which is equivalent to (x-6)²

<h3>Transformation of function</h3>

Transformation is a techniques use to change the position of an object on an xy-plane.

Given the parent function f(x) = x² and the function g(x) = x²-12x +36

Factorize g(x);

g(x) = x²-6x-6x+36

g(x)=x(x-6)-6(x-6)

Group the terms to have;

g(x) = (x-6)²

Comparing g(x) with f(x), you can see that the function f(x) is translated to the right by 6 units to produce g(x) which is equivalent to (x-6)²

Learn more on transformation here: brainly.com/question/4289712
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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
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