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Daniel [21]
3 years ago
6

Solve the equation 5x+2y=14 for y

Mathematics
1 answer:
viva [34]3 years ago
4 0

Answer:

Y could be any number under 14 because the equation does not give you enough information to answer it correctly

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Sally’s dance school had 60 students last year. This year there are only 48 students enrolled. By what percent did the enrollmen
Morgarella [4.7K]

Answer:

-20%

My brain is weird on how i figure it out but I divided 48 by 60 and got .80 so i just got the other whole to make it 1 so it is 20%. This is not the correct way to do this but this is how i got my answer.

8 0
3 years ago
What is 9m + 2 = 3m - 10
cupoosta [38]
9m+2=3m-10
subtract 3m from both sides to get:
6m+2=-10
subtract 2 from both sides to get:
6m=-12
finally, divide by 6 on both sides to get what m equals:
m=-2
7 0
3 years ago
Read 2 more answers
PLEASE PLEASE OLEASE HELP ME ASAP I WILL LOVE U FOREVER
vodomira [7]

Answer:

m=3/2

look how many units go up and to the side between 2 points

As seen here, we go up 3 units and over 2 units from (0, -3) to (2, 0.) Therefore, slope is 3/2

Step-by-step explanation:

8 0
3 years ago
2x^2+5x+3/x^2-3x-4 divided by 4x^2+2x-6/x^2-8x+16
Masja [62]

Answer: \frac{x-4}{2x-2}

Step-by-step explanation:

\frac{2x^2+5x+3}{x^2-3x-4} divided by \frac{4x^2+2x-6}{x^2-8x+16} is the same thing as multiplying \frac{2x^2+5x+3}{x^2-3x-4} by \frac{x^2-8x+16}{4x^2+2x-6}.

The equation we get is:

\frac{2x^2+5x+3}{x^2-3x-4}*\frac{x^2-8x+16}{4x^2+2x-6}

With this equation, we can factor each equation:

\frac{(x+1)(2x+3)}{(x+1)(x-4)} *\frac{(x-4)(x-4)}{2(x-1)(2x+3)}

We can cancel out like terms since they would be dividing each other:

\frac{x-4}{2x-2}

8 0
3 years ago
What factors multiply to -80 and add to 6
forsale [732]

80 = -1(80), -2(40), -4(20), -5(4),

Total answers include: -81, 79, -42, 16, -9, and 1.

None of them add up to 6!

Are you doing polynomial equations?

If so, you can solve it the hard way.

x^2 + 6x -80 = 0

(x^2 + 6x + 9) - 80 = 9

(x+3)^2 = 89

(x+3) = 9.43

x = 6.43

If you're not doing those then I'm afraid your question has no answer.

Hope that helps!



7 0
3 years ago
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