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dedylja [7]
3 years ago
12

7(x–1)–5(2x+2)=49 Please answer!

Mathematics
2 answers:
adoni [48]3 years ago
4 0

Answer:

x=-22

Step-by-step explanation:

7(x–1)–5(2x+2)=49

7x-7-10x-10=49

-3x-17=49

-3x=66

x=-22

Ivanshal [37]3 years ago
4 0

Answer:

x = -22

Step-by-step explanation:

7(x-1)-5(2x-2)=49

Step 1: Distribute

7x-7-10x-10=49

Step 2: Add like terms

-3x-17=49

Step 3: Add on both sides of the equation

-3x=66

Step 4: Divide both sides by -3 for the answer

x=-22

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Find X in the problem below <br> X=In
xxTIMURxx [149]

Answer:

50

Step-by-step explanation:

Let's drag the x line to the other side with the 14 and 48.

Then we do the Pythagorean theorem like 14^2 + 48^2 = x^2.

So then we get that x^2 is 2500

then we clearly know the square root of 2500 is 50

So 50 is our answer.

7 0
3 years ago
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Find the domain of the function f (x) = 2/x-4 + 1/x+2
Andreas93 [3]

As written, the denominator in both fractions is x, so the only restriction on the domain is ... x ≠ 0.

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We suspect you intend ...

... f(x) = 2/(x-4) +1/(x+2)

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The domain is all real numbers except -2 and 4.

6 0
3 years ago
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Harlamova29_29 [7]

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It's B

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2 times 5 = 10

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10 + 4 =14

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8 0
2 years ago
The quotient of a number 2, minus 9, is 14
rodikova [14]
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3 years ago
A rectangle is twice as long as it is wide. If its length and width are both
puteri [66]

Answer:

a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle

Step-by-step explanation:

Let l = the original length of the original rectangle

Let w = the original width of the original rectangle

From the description of the problem, we can construct the following two equations

l=2*w (Equation #1)

(l+4)*(w+4)=l*w+88 (Equation #2)

Substitute equation #1 into equation #2

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collect like terms on the same side of the equation

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4w^2+12w-72=0

Since 4 is afactor of each term, divide both sides of the equation by 4

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The quadratic equation can be factored into (w+6)*(w-3)=0

Therefore w=-6 or w=3

w=-6 can be rejected because the length of a rectangle can't be negative so

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I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.

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