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Arte-miy333 [17]
3 years ago
13

Which equation has solution x = -3? *

Mathematics
1 answer:
Alex73 [517]3 years ago
8 0

Answer:

1/2(2x-6)= -6

Step-by-step explanation:

Multiply the two sides by the equation 2

2x-6=12

move the constant to the right hand side change sign

2x = -12 + 6

calculate the sum

2x= -6

Divide the 2 sides by 2

x= -3

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Please help it’s urgent !!
schepotkina [342]

Answer:

Draw a line from (3, -2) to (3, 4).

Step-by-step explanation:

(3, -2) part: Start from 0 and go to the right 3 times, then move 2 down and draw a point there. (3, 4) part: Start from 0 and go to the right 3 times then 4 up, draw a point on that location. Connect the dots from (3, -2) to (3, 4) in a line. Then your Finished!

5 0
3 years ago
Evaluate the algebraic expression 5 + 9(x-8)2 for x = 9.<br> When x = 9,5+9(x-8)2 =
Lena [83]
23. 5+9((9)-8)2, 5+9(1)2, 5+18=23
6 0
3 years ago
X -7 &lt; -12<br><br> Solve and graph the inequality
ArbitrLikvidat [17]

Solution, x-7

Steps:

x-7

\mathrm{Add\:}7\mathrm{\:to\:both\:sides}, x-7+7

Simplify, x

The correct answer is <u><em>x<-5</em></u>

Hope this helps!!!

7 0
3 years ago
9+3.5g=11-0.5g brainliest to whoever gives steps
kow [346]

Answer:

The answer is <u>g = 0.5</u>

Step-by-step explanation:

1. Your equation is :

<em>9+3.5g=11-0.5g</em>

2. First, multiply both sides by ten

<em>9 * 10 + 3.5 * 10 = 11 * 10 - 0.5 * 10</em>

3. Next we refine the equation

<em>90 + 35g = 110 - 5g</em>

4. Now, lets subtract 90 from both sides

<em>90 + 35g - 90 = 110 - 5g - 90</em>

5. Simplify

<em>35g + 5g = -5g + 20 + 5g</em>

6. We then add 5g to both sides

<em>35g + 5g = -5g + 20 + 5g</em>

7. Simplify again

<em>40g = 20</em>

8. Now, divide both sides by 40

<em>40g/40 = 20/40</em>

9. And finally, simplify once again to get your final answer.

<em>g = 1/2 or 0.5</em>

I hope this helped :)

7 0
3 years ago
If PQ =13 and PR = 18, what is QR?
pav-90 [236]

Answer:

QR is equal to 5 since 13 is 5 less that 18. So since 13 was already taken, that leaves 5

3 0
2 years ago
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