Answer:
21
Step-by-step explanation:
Since this is a right triangle, we can use the Pythagorean theorem
a^2 +b^2 = c^2
where a and b are the legs and c is the hypotenuse (opposite the right angle)
20^2 +b^2 = 29^2
400 + b^2 =841
Subtract 400 from each side
400-400 +b^2 = 841-400
b^2 = 441
Take the square root of each side
sqrt(b^2) = sqrt(441)
b = 21
Answer:
<em>The square root of 36/196 is </em><em>3</em><em>/</em><em>7</em><em>.</em>
<em>The </em><em>square</em><em> root</em><em> of</em><em> </em><em>3</em><em>6</em><em>/</em><em>1</em><em>6</em><em>9</em><em> </em><em> </em><em>is </em><em> </em><em>6</em><em>/</em><em>1</em><em>3</em><em>.</em>
Add 't' to both sides. t+r=s
Step-by-step explanation:
It's 24k-6
By order of operations, you do the multiplication first: 3k(4) = 12k
12k-6+12k now combine like terms
24k-6
Hope that helps
From the right hand side, we will need to find a way to rewriting 3x²y in terms of cube roots.
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
![\sqrt[3]{162x^{c}y^{5}} = \sqrt[3]{(3x^{2}y)^{3}} * \sqrt[3]{6y^{d}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B162x%5E%7Bc%7Dy%5E%7B5%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%283x%5E%7B2%7Dy%29%5E%7B3%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B6y%5E%7Bd%7D%7D)
![= \sqrt[3]{162x^{6}y^{3+d}}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B162x%5E%7B6%7Dy%5E%7B3%2Bd%7D%7D)
Hence, c = 6 and d = 2