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Komok [63]
3 years ago
6

How can I solve for x?

Mathematics
2 answers:
just olya [345]3 years ago
8 0
37 is the answer, so you know that 53 is one of the angles given and there is a 90 degree angle, so all you do is 53+90=180, then you get 143=180, then subtract 180-143=37 and the real answer is 37. Hope this helps!! :)
anygoal [31]3 years ago
5 0

Answer:

37 degrees

Step-by-step explanation:

The angles within a triangle always add up to 180 degrees. Right angles are always 90 degrees. 53 + 90 = 143 degrees. 180 - 143 = 37 degrees

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Please help quickly (math)
MrMuchimi

Answer:

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3 years ago
A chemistry teacher needs to mix a 20% salt solution with a 80% salt solution to make 25 qt of a 44% salt solution. How many qua
NNADVOKAT [17]
Let there be x quarts of the 20% salt solution, and (25 - x) quarts of the 80% salt solution.
Then the total amount of salt in the 20% portion would be 0.2x, while that from the 80% portion would be (0.8)(25 - x). This would have to total up to (0.44)(25) in the final solution, so the equation is:
0.2x + 0.8(25 - x) = (0.44)(25)
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25 - x = 10 quarts of the 80% salt solution
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3 years ago
Solve for x.<br> 4 + 2x = 10
aniked [119]

Answer:

The answer is 3

Hope this helps!

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3 years ago
What is the value of (2/7)3?
igor_vitrenko [27]
Is there options for this question
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Find two rational expressions a / b and c / d that produce the result x − 1 / x2 when using the following operations. Answers
Mars2501 [29]

Answer:

a) Let \frac{a}{b}=\frac{-1}{x^2}, \text{ and } \frac{c}{d}=\frac{1}{x}.

Observe that

\frac{a}{b}+\frac{c}{d}=\frac{-1}{x^2}+\frac{1}{x}=\frac{-x+x^2}{x^3}=\frac{x(x-1)}{xx^2}=\frac{x-1}{x^2}

b)

Let \frac{a}{b}=\frac{1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x^2}.

Observe that

\frac{a}{b}-\frac{c}{d}=\frac{1}{x}-\frac{1}{x^2}=\frac{x^2-x}{x^3}=\frac{x(x-1)}{xx^2}=\frac{x-1}{x^2}

c)

Let \frac{a}{b}=\frac{x-1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x}.

Observe that

\frac{a}{b}*\frac{c}{d}=\frac{x-1}{x}*\frac{1}{x}=\frac{(x-1)1}{x*x}=\frac{x-1}{x^2}

d)

Let \frac{a}{b}=\frac{x-1}{x}, \text{ and } \frac{c}{d}=\frac{x}{1}.

Observe that

\frac{a}{b}\div\frac{c}{d}=\frac{x-1}{x}\div\frac{x}{1}=\frac{x-1}{x}*\frac{1}{x}=\frac{x-1}{x^2}

3 0
3 years ago
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