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amid [387]
3 years ago
9

-3(2x-4)-2=2(5-3x) ANSWER NEEDED ASAP please

Mathematics
1 answer:
Elan Coil [88]3 years ago
5 0

Answer:

10=10 is a true statement so x∈{R}

Step-by-step explanation:


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If the sum of two whole numbers is 24 more than their difference, then one of the number must be
yuradex [85]

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Answer would be c

Step-by-step explanation:

12+12=24   12-12=0

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Evan purchases a board game that is regularly priced at $19.95. It is on sale for 40% off. What is the sale price of the board g
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19.95 × 0.40 = $7.98.
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A textbook store sold a combined total of 480 history and math textbooks in a week. The number of history textbooks sold was two
Gwar [14]
H + m = 480
h = 2m

2m + m = 480
3m = 480
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Statistic questions
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a. 1.91

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because I think I'm lucky today

3 0
3 years ago
Find the value of the variable y, where the sum of the fraction 2/y-3 and 6/y+3 is equal to the quotient.
NISA [10]

Answer:

Here we need to solve:

\frac{2}{y - 3}  + \frac{6}{y + 3 }  = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} }

The sum of the fractions is equal to the quotient between the fractions.

Notice that the two values:

y = 3

y = -3

make the denominator equal to zero, so those values are restricted.

We can simplify the right side to get:

\frac{2}{y - 3}  + \frac{6}{y + 3 }  = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} } = \frac{2*(y + 3)}{6*(y - 3)}  = 3*\frac{y + 3}{y - 3}

Now we can multiply both sides by (y - 3)

(y - 3)*(\frac{2}{y - 3}  + \frac{6}{y + 3 }) = 3*(y + 3)\\2 + 6*\frac{y -3}{y + 3} = 3*(y + 3)

Now we can multiply both sides by (y + 3)

(2 + 6*\frac{y -3}{y + 3})*(y + 3) = 3*(y + 3)*(y + 3)

2*(y + 3) + 6*(y - 3) = 3*(y + 3)*(y + 3)\\\\2*y + 6 + 6*y - 18 = 3*(y^2 + 2*y*3 + 9)\\\\8*y - 12 = 3*y^2 + 6*y + 33\\\\0 = 3*y^2 + 6*y + 33 - 8*y + 12\\\\0 = 3*y^2 - 2*y + 45

First, let's see the determinant of that quadratic equation:

D = (-2)^2 - 4*3*45 = -536

We can see that it is negative, thus, there are no real solutions of the equation.

Thus, there is no value of y such that the origina equation is true,

6 0
3 years ago
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