a 10-by-10 grid means there are 100 squares and percent is out of 100 so if you have 42% then you just need to fill in 42 squares
8/2(2+2)=X
8/2(4)=X
4(4)=X
16=X
Your answer would be X=16
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
only works if y isn't negative.
If y could be negative, then we'd have to say
. 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.
Answer:
m<4 = 52°
m<BFD = 98°
Step-by-step explanation:
m<1 = (3x)°
m<2 = (5x - 7)°
m<3 = (4x + 15)°
m<AFD = 128°
✔️Find m<4:
m<4 = 180° - m<AFD (angles on a straight line)
Substitute
m<4 = 180° - 128°
m<4 = 52°
✔️m<BFD = m<2 + m<3
Substitute
m<BFD = (5x - 7)° + (4x + 15)°
We need to find the value of x.
Create an equation to find x.
m<1 + m<2 + m<3 = m<AFD (angle addition postulate)
Substitute
3x + 5x - 7 + 4x + 15 = 128°
Add like terms and solve for x
12x + 8 = 128
12x + 8 - 8 = 128 - 8
12x = 120
12x/12 = 120/12
x = 10
m<BFD = (5x - 7)° + (4x + 15)°
Plug in the value of x
m<BFD = 5(10) - 7 + 4(10) + 15
m<BFD = 50 - 7 + 40 + 15
m<BFD = 98°
We have to find the expansion of 
We will use binomial expansion to expand the given expression, which states that the expression
is expanded as :

Now expanding
we get,


So, the variables are
,
,
, ![a^{8} , [tex] ab^{7}](https://tex.z-dn.net/?f=%20a%5E%7B8%7D%20%20%2C%20%5Btex%5D%20ab%5E%7B7%7D%20)