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Paraphin [41]
3 years ago
8

What adds to get 3 but multiplies to get negative 18

Mathematics
1 answer:
Colt1911 [192]3 years ago
3 0
6 and -3

6 + -3 = 3

6  x -3 = -18

Hope that helps! 
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What is an approximate average rate of change of the graph from x = 12 to x = 15, and what does this rate represent? (4 points)
nirvana33 [79]

Answer: first of all, i would put it in an equation. then i would slove that and use that answer.


Step-by-step explanation:  

use the equation <u>12X x 15 = y </u>

<u>12x=15</u>

12     12

x= 180

4 0
4 years ago
How do I write an algebraic equation for this word problem?
Doss [256]
Hi.

Please note that I haven't done much algebra.

4 more = 4 +

Twice as many than the runner up means 2 times the amount they ate (a variable) = 2n

So far we have 4 + 2n

Since it's an equation, we are going to know what it equals. In the problem, we find out that it's 11. So the equation looks like this:

4 + 2n = 11

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7 0
3 years ago
Which function increases by 3 every time the x-increases by 1? A) y = 3x B) y = 3x + 1 C) y = 1 3 x D) y = 1 3 x + 1
dexar [7]

Answer:

B

Step-by-step explanation:

as it says 3 everytime the x increase by 1

4 0
3 years ago
Read 2 more answers
What is the answer to this math problem
shutvik [7]
Y + x = 5x + 3
12 - y = x + 2y

Isolate x in the first equation, then substitute its expression into the second equation to find y:

y + x = 5x + 3
x = 5x - y + 3
-4x = -y + 3
x = y - 3/4

---

12 - y = x + 2y
12 - y = (y - 3/4) + 2y
48 - 4y = y - 3 + 8y
48 - 4y = 9y - 3
-4y = 9y -51
-13y = -51
y = 51/13

Substitute this into the first equation to find x:

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x - 5x = 3 - 51/13
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x = 3/13

Lastly, substitute back into both original equations to check work;

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12 - y = x + 2y   -->   12 - 51/13 = 3/13 + 2(51/13)  -->  105/13 = 105/13

Answer:
x = 3/13
y = 51/13

Hope this helps.
4 0
3 years ago
A moving truck can carry up to 1,325 pounds. The truck can carry 25 of the same-sized boxes. If each box weighs the same amount,
anastassius [24]

Answer:

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Step-by-step explanation:

5 0
4 years ago
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