Answer:
Option D: 1/3
Step-by-step explanation:
Let
x ------> the entire Stefanie's allowance
we know that
She spends on clothes (1/2)x
She spends on movies (1/6)x
To find out how much more does she spend on clothes that he does on movies, subtract the amount she spends on movies from the amount she spend on clothes
Simplify
therefore
She spends 1/3 more on clothes that he does on movies
Answer will be (D)
The value of the exponential function will eventually be greater than the
linear function.
and
B)
The value of the linear function will never be greater than the exponential
function
Answer:
![\huge \purple {\frac{2 \sqrt[3]{6} + \sqrt[3]{18} }{ 6} }](https://tex.z-dn.net/?f=%20%5Chuge%20%20%5Cpurple%20%7B%5Cfrac%7B2%20%5Csqrt%5B3%5D%7B6%7D%20%2B%20%20%5Csqrt%5B3%5D%7B18%7D%20%7D%7B%20%206%7D%20%20%7D)
Step-by-step Explanation:
![\huge \frac{2 + \sqrt[3]{3} }{ \sqrt[3]{6} } \\ \\ = \huge \frac{(2 + \sqrt[3]{3} )}{ \sqrt[3]{6} } \times \frac{ \sqrt[3]{6} \times \sqrt[3]{6} }{\sqrt[3]{6} \times \sqrt[3]{6}} \\ \\ = \huge \frac{(2 + \sqrt[3]{3} )}{ \sqrt[3]{6} } \times \frac{ \sqrt[3]{6 ^{2} } }{\sqrt[3]{6^{2}} } \\ \\ = \huge \frac{(2 + \sqrt[3]{3} )\sqrt[3]{6} }{ \sqrt[3]{6} \times \sqrt[3]{6 ^{2} }} \\ \\ = \huge \frac{(2 \sqrt[3]{6} + \sqrt[3]{3} \sqrt[3]{6} )}{ \sqrt[3]{6 ^{3} }} \\ \\ = \huge \orange{\frac{2 \sqrt[3]{6} + \sqrt[3]{18} }{ 6} }](https://tex.z-dn.net/?f=%20%5Chuge%20%5Cfrac%7B2%20%2B%20%20%5Csqrt%5B3%5D%7B3%7D%20%7D%7B%20%5Csqrt%5B3%5D%7B6%7D%20%7D%20%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%5Chuge%20%5Cfrac%7B%282%20%2B%20%20%5Csqrt%5B3%5D%7B3%7D%20%29%7D%7B%20%5Csqrt%5B3%5D%7B6%7D%20%7D%20%20%5Ctimes%20%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7B6%7D%20%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B6%7D%20%7D%7B%5Csqrt%5B3%5D%7B6%7D%20%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B6%7D%7D%20%20%5C%5C%20%20%5C%5C%20%3D%20%20%5Chuge%20%5Cfrac%7B%282%20%2B%20%20%5Csqrt%5B3%5D%7B3%7D%20%29%7D%7B%20%5Csqrt%5B3%5D%7B6%7D%20%7D%20%20%5Ctimes%20%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7B6%20%5E%7B2%7D%20%7D%20%20%7D%7B%5Csqrt%5B3%5D%7B6%5E%7B2%7D%7D%20%20%7D%20%5C%5C%20%20%5C%5C%20%3D%20%20%5Chuge%20%5Cfrac%7B%282%20%2B%20%20%5Csqrt%5B3%5D%7B3%7D%20%29%5Csqrt%5B3%5D%7B6%7D%20%7D%7B%20%5Csqrt%5B3%5D%7B6%7D%20%5Ctimes%20%5Csqrt%5B3%5D%7B6%20%5E%7B2%7D%20%7D%7D%20%20%5C%5C%20%20%5C%5C%20%3D%20%20%5Chuge%20%5Cfrac%7B%282%20%5Csqrt%5B3%5D%7B6%7D%20%2B%20%20%5Csqrt%5B3%5D%7B3%7D%20%5Csqrt%5B3%5D%7B6%7D%20%29%7D%7B%20%20%5Csqrt%5B3%5D%7B6%20%5E%7B3%7D%20%7D%7D%20%20%5C%5C%20%20%5C%5C%20%3D%20%20%5Chuge%20%20%5Corange%7B%5Cfrac%7B2%20%5Csqrt%5B3%5D%7B6%7D%20%2B%20%20%5Csqrt%5B3%5D%7B18%7D%20%7D%7B%20%206%7D%20%20%7D)
By multiplying the factors together you’ll get 6
One application of volume is determining the density of an object. Assume the object is made of a pure element (eg: gold). If we know the volume (v) of the object, and we know the mass (m), then we can use the formula D = m/v to figure out the density D. Knowing the volume is also handy to determine if the object can fit into a larger space or not. Another application is figuring out how much water is needed to fill up the inner space of the 3D solid (assuming it's hollow on the inside).
The surface area is handy to figure out how much material is needed to cover the outer surface. This material can be paint, paper, metal sheets, or whatever you can think of really. A good example is wrapping a present and the assumption is that there is no overlap.
