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kati45 [8]
3 years ago
6

20 POINTS NEED HELP ASAP QUESTION IN PICTURE BELOW

Mathematics
1 answer:
crimeas [40]3 years ago
8 0
For this case we have the following solution.
 x = gallons of water to be added
 For the 10% solution we have:
 0.1 * 8 = 0.8
 Then, for 5% we have:
 (0.8 / x + 8) = 0.05
 Rewriting:
 (0.8 / x + 8) = (5/100)
 Answer:
 An equation can be used to find x, the humber of gallons of water he should add is:
 (0.8 / x + 8) = (5/100) 
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Step-by-step explanation:

for perimeter, you add x+x+x+x+1+1, which is 4x+2

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Which expression is equivalent to square root 55x^7y6/11x^11y^8
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Answer:

\large\boxed{\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\dfrac{\sqrt5}{x^2y}}

Step-by-step explanation:

\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{55}{11}\cdot\dfrac{x^7}{x^{11}}\cdot\dfrac{y^6}{y^8}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=\sqrt{5x^{7-11}y^{6-8}}=\sqrt{5x^{-4}y^{-2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt5\cdot\sqrt{x^{-4}}\cdot\sqrt{y^{-2}}=\sqrt5\cdot\sqrt{x^{(-2)(2)}}\cdot\sqrt{y^{(-1)(2)}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt5\cdot\sqrt{(x^{-2})^2}\cdot\sqrt{(y^{-1})^2}\qquad\text{use}\ \sqrt{a^2}=a

=\sqrt5\cdot x^{-2}\cdot y^{-1}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\to a^{-1}=\dfrac{1}{a}\\\\=\sqrt5\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}=\dfrac{\sqrt5}{x^2y}

6 0
3 years ago
What happens when an image is reflected over y = -x
notka56 [123]

Answer:

Well, it's pretty simple. If your reflecting something over the y axis the X will change if it's negative or positive. So if it's -x and you flip it over it becomes a positive x. If you flip it over the x-axis the Y is the one that change to a negative or positive.

So if the shape flips over the y-axis, the X points will turn negative. for example, one of the points is (1,4) it will turn to (-1,4)

Step-by-step explanation:


7 0
3 years ago
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In-s [12.5K]
VW=CD since they are congruent figure, so CD=6
7 0
3 years ago
Read 2 more answers
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