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erastova [34]
3 years ago
9

PLEASE HELP! I NEED THIS

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
6 0

Answer:

x² + x - 30

Step-by-step explanation:

Given

(x - 5)(x + 6)

Each term in the second factor is multiplied by each term in the first factor, that is

x(x + 6) - 5(x + 6) ← distribute both parenthesis

x² + 6x - 5x - 30 ← collect like terms

= x² + x - 30

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Need some help!<br> please be specific on how to do it
Katyanochek1 [597]

Answer:

Your first step in solving the system would be to substitute the value of a variable into one of the equations

Step-by-step explanation:

We know that x = 2y + 2 and y = -1/3x + 4, so we could insert one of these values as the variable in one of the equations

If we were to insert -1/3x + 4 for y, our new equation would be:

x = 2(-1/3x + 4) + 2 (as you can see, the value of y was inserted into the equation)

4 0
2 years ago
Multiples that are shared by two or more numbers are_______
Elanso [62]

Answer:

it is called Common Multiples

3 0
3 years ago
Read 2 more answers
Use an algebraic approach to solve the problem.
qaws [65]

Answer:

Her first exam score was a 78, her second exam score was a 89, and his third exam score was a 94.

Step-by-step explanation:

Use the formula for the mean: sum of elements / number of elements

Let x represent her first exam score.

Her second exam score can be represented by x + 11, since it was 11 points better than her first.

Her third exam score can be represented by (x + 11) + 5, since it was 5 points better than her second.

Plug in all of these expressions into the mean formula. Plug in 87 as the mean, and plug in 3 as the number of elements (since there are 3 scores):

mean = sum of elements / number of elements

87 =  ( (x) + (x + 11) + (x + 11) + 5 ) / 3

Add like terms and solve for x:

87 = (3x + 27) / 3

261 = 3x + 27

234 = 3x

78 = x

So, her first score was a 78.

Find her second score by adding 11 to this:

78 + 11 = 89

Find her third score by adding 5 to the second score:

89 + 5 = 94

Her first exam score was a 78, her second exam score was a 89, and his third exam score was a 94.

6 0
3 years ago
What is the following product? (2 square root 7+3 square root 6)(5 square root 2+4 square root 3)
Xelga [282]

(2\sqrt7+3\sqrt6)(5\sqrt2+4\sqrt3)\qquad\text{use distributive property}\\\\=(2\sqrt7)(5\sqrt2)+(2\sqrt7)(4\sqrt3)+(3\sqrt6)(5\sqrt2)+(3\sqrt6)(4\sqrt3)\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt{12}+12\sqrt{18}\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt{4\cdot3}+12\sqrt{9\cdot2}\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt4\cdot\sqrt3+12\sqrt9\cdot\sqrt2\\\\=10\sqrt{14}+8\sqrt{21}+(15)(2)\sqrt3+(12)(3)\sqrt2\\\\=\boxed{10\sqrt{14}+8\sqrt{21}+30\sqrt3+36\sqrt2}

5 0
3 years ago
Read 2 more answers
Repost of my question i need help quicklyy help plzz
GuDViN [60]

Answer:

According to logarithmic properties.... The right hand side can be written as

log base 7 (180/3).....which is log base 7 60

So according to the question cancel out the log base 7 from both sides....

Then we get 8r + 20 = 60

That is 8r = 40..

That is r = 5......

Therefore the value of r is 5

8 0
3 years ago
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