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natka813 [3]
1 year ago
10

Margie's work for adding linear expressions is shown below. After checking her answer with the answer key, she solved it incorre

ctly. Given (−1.56b + 10) − (4.27b − 14) Step 1 −1.56b + 10 + (−4.27b) + 14 Step 2 −1.56b + 4.27b + 10 + 14 Step 3 (−1.56b + 4.27b) + (10 + 14) Step 4 2.71b + 24 Part A: Identify and explain the first step where Margie made an error. (6 points) Part B: Explain how to correctly write the expression in fewest terms by correcting the error in Part A. Show all work. (6 points)
Mathematics
1 answer:
xxMikexx [17]1 year ago
5 0

The correct form of expression after solving the errors is :

-5.83b + 24

Given Margie's work for adding linear expressions are:

the expression is (−1.56b + 10) − (4.27b − 14)

step 1: −1.56b + 10 + (−4.27b) + 14

In the first step of margie, she did not used the proper method of opening the brackets.

The negative sign is used to multiply the terms inside the brackets.

so the correct step is:

step 1: -1.56b + 10 - 4.27b + 14

in the next step arrange the variables and constants.

step 2 :  -1.56b - 4.27b +10 + 14

next add the variables and the constants:

step 3: (-1.56b - 4.27b) +(10 + 14)

step 4: -5.83b + 24

Hence we get the required results.

Learn more about Solving expressions here:

brainly.com/question/723406

#SPJ1

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Use the quadratic formula to solve the following equation -3x^2-x-3=0
tigry1 [53]

<u>Answer:</u>

x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i are two roots of equation -3 x^{2}-x-3=0

<u>Solution:</u>

Need to solve given equation using quadratic formula.

-3 x^{2}-x-3=0

General form of quadratic equation is a x^{2}+b x+c=0

And quadratic formula for getting roots of quadratic equation is

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}

Since b^{2}-4 a c is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}

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