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enyata [817]
4 years ago
7

Correct the error a student made when solving the equation 4=-2(x-3). What is the correct solution?​

Mathematics
2 answers:
Murrr4er [49]4 years ago
8 0
In the first step when the student did -2x-3 they didn’t make the sign positive. It should be 4=-2+6, not 4=-2-6
tankabanditka [31]4 years ago
3 0

Answer:

It should be positive 6

Step-by-step explanation:

The student multiplied a negative and a negative forgetting that it would become a positive. So in this case x would equal 1

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Simplify the expression:<br> 6(h + 6) + 1
Brrunno [24]

Answer:

This is your answer

4 0
3 years ago
Simplify and round the expression below. C isn’t necessary to answer
myrzilka [38]
11.3057558 is the answer when you multiply/add/subtract it all.
6 0
3 years ago
Can someone help me please
Schach [20]

Answer:

Step-by-step explanation:

The system of equations is expressed as

y=-2x+7 - - - - - - - - 1

y=5x-7 - - - - - - - - -2

We would equate equation 1 and equation 2. It becomes

- 2x + 7 = 5x - 7

We would add 2x to the left hand side and the right hand side of the equation. Also add 7 to the left hand side and the right hand side of the equation. It becomes

- 2x + 2x + 7 + 7 = 5x + 2x - 7 + 7

14 = 7x

Dividing the left hand side and the right hand side of the equation by 7, it becomes

7x/7 = 14/2

x = 2

y= -2 × 2 + 7

y = - 4 + 7

y = 3

7 0
4 years ago
8g-14g+22=37 what does g equal
Sphinxa [80]
If you would like to solve the equation 8 * g - 14 * g + 22 = 37 for g, you can calculate this using the following steps:

8 * g - 14 * g + 22 = 37
- 6 * g = 37 - 22
- 6 * g = 15   /(-6)
g = 15 / (-6)
g = - 15 / 6
g = - 5 / 2 = - 2.5

Result: g equals to - 5/2.
5 0
4 years ago
Read 2 more answers
What are the solutions of 3x2 + 6x + 6 = 0?
AlladinOne [14]
The solutions or roots to this equation is found by solving for x. We can do this a couple ways either by FOIL or quad formula

3x^2+6x+6=0
3(x^2+2x+2)=0

x^2+2x+2=0 we cant FOIL this out so we use the quad formula

x= [(-b+\-sqrt(b^2-4ac))/2a]

x= -2+\- sqrt(4-4(1)(2))/2(1)
x= -1 +i , -1 - i

So we have complex roots since our quad formula returned a negative number. Whenever the quad formula answer is positive we have two roots/solutions, when it is zero we have one root/solution, and whenever it is negative we have Complex roots/solutions

Hope this helps. Any questions please just ask. Thank you.
7 0
3 years ago
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