Answer:
The equivalent expression for the given expression ![\sqrt[3]{256x^{10}y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D)
is
![4x^{3} y^{2}(\sqrt[3]{4xy} )](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%20%29)
Step-by-step explanation:
Given:
Solution:
We will see first what is Cube rooting.
![\sqrt[3]{x^{3}} = x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D%20%3D%20x)
Law of Indices
Now, applying above property we get
![\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D%3D%5Csqrt%5B3%5D%7B%284%5E%7B3%7D%5Ctimes%204%5Ctimes%20%28x%5E%7B3%7D%29%5E%7B3%7D%5Ctimes%20x%5Ctimes%20%28y%5E%7B2%7D%29%5E%7B3%7D%5Ctimes%20y%20%20%20%29%7D%20%5C%5C%5C%5C%5Ctextrm%7BCube%20Rooting%20we%20get%7D%5C%5C%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D%3D%204%5Ctimes%20x%5E%7B3%7D%5Ctimes%20y%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%29%20%5C%5C%5C%5C%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D%3D%204x%5E%7B3%7Dy%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%29)
∴ The equivalent expression for the given expression is
For the first one, we know that is a right angle. right angles are 90 degrees. if we subtract 90 - 49 that equals 41. so the second value needs to equal 41. since we have a 3 there already, we are going to subtract 41-3, which is 38. x = 38
The bisector of the angle at A (call it AQ) divides the segment BC into segments BQ:QC having the ratio AB:AC. Use this fact to find x.
.. 9:15 = (2x -1):3x
.. 15(2x -1) = 9*3x . . . . . the product of the means equals the product of extremes
.. 30x -15 = 27x
.. 3x = 15
.. x = 5
___
According to the value of x, the bisector AQ divides the triangle into two isosceles triangles: ABQ, ACQ.
Answer:
Your final answer is either
x≥-2 if your initial inequality was
6x+2≤2(5-x)
OR
x≤-2
if your initial inequality was
6x+2≥2(x-2)
Step-by-step explanation:
As shown you have an equality, not an inequality.
-6x+2=2(5-x) distribute through parenthesis
-6x+2=2(5)+2(-x)
-6x+2=10-2x add 2x to both sides
2x-6x+2=10-2x+2x
-4x+2=10 subtract 2 from both sides
-4x+2-2=10-2
-4x=8 divide both sides by -4
-4x/(-4) = 8/(-4)
x = -2
With the ≥ or ≤ sign you would solve the exact same way
except for the point where when dividing both sides by
-4 requires you to reverse the inequality.
Your final answer is either
x≥-2 if your initial inequality was
6x+2≤2(5-x)
OR
x≤-2
if your initial inequality was
6x+2≥2(x-2)
Answer:
Alright well solve for the variable in one of the equation's. then substitute the result of the other equation
Point form: - 175/51, 1/51
Equation form:
x = - 175/51 y = 1/51 Hope this help's :)
Step-by-step explanation:
