Answer:
1/3(40)ft³
Step-by-step explanation:
volume of cone=1/3 volume of cylinder
hope this helps
plzz mark me brainliest
Answer:
The answer is y = 4x - 1.
Step-by-step explanation:
☆The equation we'll be using: y = mx + b
☆Since we already have the slope. we have to find the y-intercept.
☆
☆Using the information we gathered and was given, we just to put them together.
☆
Answer:
Surface area = 663π in².
Volume = (676/3)π in² ≈ 225.33 π in²
Explanation:
1) We know the radius and the lateral area.
2) With the radius you can find the areas of the top and the bottom.
For that, you use the formula:
area of the top = area of the bottom = π r²
∴ π (13 in)² = 169π in² (each)
3) Then, the surface area is the sum of the lateral area and the two bases (top and bottom)
surface area = lateral area + bottom area + top area = 325π in² + 2×169π in² = 663π in².
3) You can also find the height of the cylinder.
Use the formula: lateral area = 2π r h
∴ h = lateral area / [2 π r]
⇒ h = 325 π / [ 2π (13) ] = 12.5 in
4) With the height you can find the volume.
Use the formula: V = (4/3) π r³
∴ V = (4/3) π (13 in)³ = (676/3)π in² ≈ 225.33 π in²
![\bf \cfrac{\sqrt[4]{63}}{4\sqrt[4]{6}}\qquad \begin{cases} 63=3\cdot 3\cdot 7\\ 6=2\cdot 3 \end{cases}\implies \cfrac{\sqrt[4]{3\cdot 3\cdot 7}}{4\sqrt[4]{2\cdot 3}}\implies \cfrac{\underline{\sqrt[4]{3}}\cdot \sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}\cdot \underline{\sqrt[4]{3}}} \\\\\\ \cfrac{\sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{3\cdot 7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B63%7D%7D%7B4%5Csqrt%5B4%5D%7B6%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0A63%3D3%5Ccdot%203%5Ccdot%207%5C%5C%0A6%3D2%5Ccdot%203%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%5Ccdot%203%5Ccdot%207%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5Ccdot%203%7D%7D%5Cimplies%20%5Ccfrac%7B%5Cunderline%7B%5Csqrt%5B4%5D%7B3%7D%7D%5Ccdot%20%5Csqrt%5B4%5D%7B3%7D%5Ccdot%20%5Csqrt%5B4%5D%7B7%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%5Ccdot%20%5Cunderline%7B%5Csqrt%5B4%5D%7B3%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%7D%5Ccdot%20%5Csqrt%5B4%5D%7B7%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%5Ccdot%207%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D)
![\bf \textit{now, rationalizing the denominator}\\\\ \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}\cdot \cfrac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21}\cdot \sqrt[4]{8}}{4\sqrt[4]{2}\cdot \sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21\cdot 8}}{4\sqrt[4]{2\cdot 2^3}}\implies \cfrac{\sqrt[4]{168}}{4\sqrt[4]{2^4}} \\\\\\ \cfrac{\sqrt[4]{168}}{4\cdot 2}\implies \cfrac{\sqrt[4]{168}}{8}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bnow%2C%20rationalizing%20the%20denominator%7D%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B2%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B2%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%5Ccdot%20%5Csqrt%5B4%5D%7B8%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%5Ccdot%20%5Csqrt%5B4%5D%7B2%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%5Ccdot%208%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5Ccdot%202%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5E4%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B4%5Ccdot%202%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B8%7D)
and is all you can simplify from it.
so... all we did, was rationaliize it, namely, "getting rid of the pesky radical at the bottom", we do so by simply multiplying it by something that will raise the radicand, to the same degree as the root, thus the radicand comes out.
Answer:
bueh
Step-by-step explanation:
