SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

Mathematics

1 answer:

svet-max [94.6K]3 years ago

3 0

Remember to be mindful of signs when adding and subtracting. 2c^2+5c+4

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3 = 2 – x 3 = 2 – 2 – x 3 = –x 2003-03-02-00-00_files/i0110000.jpg –3 = x A. The student incorrectly applied the multiplication

rewona [7]

Answer:
B. The student did not properly apply the addition property to isolate x

Explanation:
When given an equation to solve, always remember that when you do an external operation (add/subtract/multiply a term or divide by a term) on one side of the equation, the same operation should be applied on the other side in order to maintain the equality of the equation.

Now, let's take a look on the steps done:
Step 1:
3 = 2 - x
Step 2:
3 = 2 - 2 - x
Step 3:
3 = -x

Now, note n step 2, the student wanted to get rid of the 2 next to the x, therefore, he subtracted 2. However, the student did not subtract the 2 from the other side of the equation. Since we're taking addition (we're adding a -2), therefore, the student incorrectly applied the addition property to isolate the x.

The correct steps would be as follows:
Step 1:
3 = 2 - x
Step 2:
3 - 2 = 2 - 2 - x
Step 3:
1 = - x

Hope this helps :)

5 0

3 years ago

Use this image to classify each pair of lines.

irga5000 [103]

Answer:

Parallel

Step-by-step explanation:

7 0

1 year ago

Given a quadratic function in vertex form:

Alja [10]

Answer:

See below

Step-by-step explanation:

Part A

A vertical stretch takes place when $a\cdot f(x)$ given that $|a| > 1$

Part B

A vertical compression takes place when $a\cdot f(x)$ given that $0 < |a| < 1$

Part C

A vertical stretch is different than a horizontal compression:

In a vertical stretch, the input stays the same, but the output is multiplied by the scale factor
In a horizontal compression, the output stays the same, but the input is multiplied by the scale factor

Part D

A reflection across the x-axis means that the output, our y-variable, is the opposite sign. This means that all values of $a$ must be negative such that $a\cdot f(x)$ as mentioned in parts A and B. Also, in part C, since our scale factor is negative, the output is the only one being multiplied by the scale factor.