Answer:
Step-by-step explanation:
First Question
The top question has 2 solutions because eventually you will have to take the square root of (x - 4)^2. One square root will be plus, and the other minus
(x - 4)^2 - 28 = 8 Add 28 to both sides
(x - 4)^2 -28+28 = 8+28 Collect like terms
(x - 4)^2 = 36 Take the square root.
sqrt( (x - 4)^2 ) = sqrt(36)
x - 4 = +/- 6
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x - 4 = + 6
x = 6 + 4
x = 10
============
x - 4 = - 6
x = - 6 + 4
x = - 2
The two solutions are (10,0) and (-2,0)
Second Question
The question is solved correctly up to the point where the square root needs to be taken. It is incorrect at that point.
x^2 - 6x - 7 = 0 Add 7 to both sides.
x^2 - 6x = 7 Add 1/2 the linear term and square it to both sides.
x^2 - 6x + (6/2)^2 = 7 + (6/2)^2 Expand the right
x^2 - 6x + 3^2 = 7 + 9
x^2 - 6x + 9 = 16 express the square on the left
(x - 3)^2 = 16 Take the sqrt of both sides.
x - 3 = sqrt(16)
x - 3 = +/-4
============
x - 3 = 4
x = 4 + 3
x = 7
===========
x -3 = - 4
x = - 4 + 3
x = - 1
==========
Solutions
(7,0) and (-1,0)
Answer: 15 cookies a weekday
Step-by-step explanation:
22 - 7 = 15
this is because we exclude the cookies he eats on weekends because that is not a week day so we are left with 15
Answer:
B
Step-by-step explanation:
BECAUSEB UR DFGHHGDUHGJDHGJV GRJFDHHFJDG GHNHHD
<u>Corrected Question</u>
If f(x) = -3x+4 and g(x) = 2, solve for the value of x for which f(x) = g(x) is true.
Answer:
![x=\dfrac{2}{3}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B2%7D%7B3%7D)
Step-by-step explanation:
Given the functions:
When f(x)=g(x), we have:
-3x+4=2
Collect like terms by subtracting 4 from both sides
-3x+4-4=2-4
-3x=-2
Divide both sides by -3 to solve for x.
![\dfrac{-3x}{-3}=\dfrac{-2}{-3}\\ x=\dfrac{2}{3}\\$Therefore, $f(\dfrac{2}{3})=g(\dfrac{2}{3})](https://tex.z-dn.net/?f=%5Cdfrac%7B-3x%7D%7B-3%7D%3D%5Cdfrac%7B-2%7D%7B-3%7D%5C%5C%20%20x%3D%5Cdfrac%7B2%7D%7B3%7D%5C%5C%24Therefore%2C%20%24f%28%5Cdfrac%7B2%7D%7B3%7D%29%3Dg%28%5Cdfrac%7B2%7D%7B3%7D%29)
We conclude therefore that at
, the values of f(x) and g(x) are equal.