Answer:
Divide the 1st equation by 2
a + 2c = 19
Divide the 2nd equation by 3
a + c = 13.5
Subtract the 1st equation and the 2nd equation
c = 5.5
Substituting back into the 2nd equation will give:
a + 5.5 = 13.5
a = 8
Step-by-step explanation:
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
I believe the answer is A because the square root of 10 root 3 is approximately 17.32.
Answer:
Volume of a parallelogram is v=whl
Area of s parallelogram is A= bh
Circumference of a circle= rd
Area of a circle= rd^2
Area of a triangle is 1/a <em>bh</em>
Area of a trapezoid is 1/2(b1 + b2)h
Perimeter of a quadrilateral is 2L + 2W
I couldn't drag the image but I knew what I was doing!
