Answer:
y=5 when x=20
Step-by-step explanation:
Using the format for inverse proportions: y=k/x
To find k, substitute 25 in for y and 4 for x to find:
25=k/4 ⇒ k=100
Using this we can substitute k and x in for 100 and 20 to find:
y=100/20 ⇒ y=5
First, you want to solve for the equation for this problem, would be:
0.05N + 0.10D = 20.50
While N = The amount of nickels, and D = The amount of dimes.
Since N = 164, and D = 123. It would add up to 287 coins. 164 + 123 = 287.
Now that you have the number for those two variables, solve the equation for when N = 164, and D = 123.
0.05N + 0.10D = 20.50
<span>0.05(164) + 0.10(123) = 20.50
</span><span>8.2 + 12.3 = 20.50
</span>20.50 = 20.50
So, there is 164 nickels, and 123 dimes for your answer.
<em>I hope this helps! </em>
<em>~ Notorious Sovereign</em>
Answer: B
Step-by-step explanation:
Oki so I know this becaue I study to much and because if there was an answer saying this : y=6x+3 it would be wrong because it is adding to the constant
So y=6x is proportinal because it is allways constant. I forgot this was an example sorry.
I think that the answer to the question is b
Answer: ![\frac{\sqrt[4]{10xy^3}}{2y}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D)
where y is positive.
The 2y in the denominator is not inside the fourth root
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Work Shown:
![\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%7D%7B8y%7D%7D%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%2A2y%5E3%7D%7B8y%2A2y%5E3%7D%7D%5C%20%5C%20%5Ctext%7B....%20multiply%20top%20and%20bottom%20by%20%7D%202y%5E3%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B10xy%5E3%7D%7B16y%5E4%7D%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B16y%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20break%20up%20the%20fourth%20root%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20rewrite%20%7D%2016y%5E4%20%5Ctext%7B%20as%20%7D%20%282y%29%5E4%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D%20%5C%20%5C%20%5Ctext%7B...%20where%20y%20is%20positive%7D%5C%5C%5C%5C%5C%5C)
The idea is to get something of the form 
in the denominator. In this case, 
To be able to reach the 
, your teacher gave the hint to multiply top and bottom by 
For more examples, search out "rationalizing the denominator".
Keep in mind that ![\sqrt[4]{(2y)^4} = 2y](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%20%3D%202y)
only works if y isn't negative.
If y could be negative, then we'd have to say ![\sqrt[4]{(2y)^4} = |2y|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%20%3D%20%7C2y%7C)
. The absolute value bars ensure the result is never negative.
Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.
