Make a substitution:
Then the system becomes
![\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu-v%7D%2B%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu-v%7D-%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
Simplifying the equations gives
![\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
which is to say,
![\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%5Ctimes4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D)
![\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81](https://tex.z-dn.net/?f=%5Cimplies%5Csqrt%5B3%5D%7B%5Cleft%28%5Cdfrac%20uv%5Cright%29%5E4%7D%3D81)
Substituting this into the new system gives
![\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7B%28%5Cpm27v%29%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cimplies%5Cdfrac1%7Bv%5E2%7D%3D1%5Cimplies%20v%3D%5Cpm1)
Then
(meaning two solutions are (7, 13) and (-7, -13))
Answer:
The answer is 1.62 ounce
Step-by-step explanation:
<h3>
<u>Given</u>;</h3>
- Jaydon has eaten 3.6% of his 45 ounce block of chocolate fudge.
<h3>
<u>T</u><u>o</u><u> Find</u>;</h3>
- How many ounces of fudge has he eaten?
Now,
3.6% of his 45 ounce
45 × 3.6 ÷ 100
162 ÷ 100 = 1.62
Thus, Jaydon has eaten 1.62 ounce block of chocolate fudge.
Answer:
E.200.5ft²
Step-by-step explanation:
To find the area of the composite figure we simply split it into two regular shapes, a half circle and a square, we then find the area of the two shapes individually and add them together.
Area of square = s² where s = side length
It appears the given side length is 12 so s = 12
Which means area = 12² = 144ft²
For semi circle
Area = 1/2(πr² ), where r is the radius
The side length of the square is shared with the diameter of the semi circle meaning that the diameter of the semi circle is 12
To convert to radius from diameter we simply divide by 2 so r = 12/2 = 6
We have area = 1/2(πr² ) and r = 6
So area = 1/2(π6²)
==> evaluate exponent
Area = 1/2(36π)
==> take one half of 36
Area = 18π
==> multiply 18 and π
Area = about 56.5
Finally we add the two areas together
Total area = 56.5 + 144 = 200.5ft²
1 yard = 3 feet
Divide 34 by 3, you should get 11 and 1/3.
So the classroom is 11 1/3 yards long, but 1/3 of a yard just so happens to be one foot, as 3 feet is 1 yard.
That means that the classroom is 11 yards and 1 foot long, or answer choice B.
Hope this helps! :)
Yes. 1 x 6 = 6
2 x 3 = 6
3 x 2 = 6
6 x 1 = 6
