Answer:
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Answer
Is there any numbers or is it just like that?
Answer: 0=0
Step-by-step explanation:
First you want to solve for x
2x+y=−4
Step 1: Add -y to both sides.
2x + y + −y =−4+ −y
2x=−y−4
Step 2: Divide both sides by 2.
2x/2= -y-4/2
x=-1/2y-2
Then you want to plug in your answer for x into the equation so:
2(-1/2y-2)+y=-4
Step 1: Simplify both sides of the equation.
2(-1/2y-2)+y=-4
(2)(
−1/
2
y)+(2)(−2)+y(Distribute)
−y+−4+y=−4
(−y+y)+(−4)=−4(Combine Like Terms)
−4=−4
−4=−4
Step 2: Add 4 to both sides.
−4+4=−4+4
0=0.
Sorry if this wasn't the answer your looking for. If you need more help I suggest using Ma.th.Pa.pa calculator but with out the periods. Hope you have a good day :)
Answer:
1. ![(\sqrt[5]{(m+2)})^{3} = (m+2)^{\frac{3}{5}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B5%5D%7B%28m%2B2%29%7D%29%5E%7B3%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D)
2. ![(\sqrt[3]{(m+2)})^{5} = (m+2)^{\frac{5}{3}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B%28m%2B2%29%7D%29%5E%7B5%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D)
3. ![\sqrt[5]{(m)}^{3}+2 = m^{\frac{3}{5}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%28m%29%7D%5E%7B3%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D%2B2)
4. ![\sqrt[3]{(m)}^{5}+2 = m^{\frac{5}{3}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%28m%29%7D%5E%7B5%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%2B2)
Step-by-step explanation:
Recall that
![(\sqrt[n]{x})^{m} = (x^{\frac{m}{n}})](https://tex.z-dn.net/?f=%28%5Csqrt%5Bn%5D%7Bx%7D%29%5E%7Bm%7D%20%3D%20%20%28x%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%29)
Where 
is called radicand and n is called index
1. Root(5, (m + 2) ^ 3)
In this case,
n is 5
m is 3
x = (m + 2)
2. Root(3, (m + 2) ^ 5)
In this case,
n is 3
m is 5
x = (m + 2)
3. Root(5, m ^ 3) + 2
In this case,
n is 5
m is 3
x = m
4. Root(3, m ^ 5) + 2
In this case,
n is 3
m is 5
x = m
Answer:
AB ≈ 35.93
Step-by-step explanation:
Using the cosine ratio in the right triangle
cos3.5° = 
= 
= 
Multiply both sides by 36
36 × cos3.5° = AB, thus
AB ≈ 35.93 ( to 2 dec. places)
