Answer:
x > 2 is your answer.
Step-by-step explanation:
Isolate the variable, x. Treat the > sign as an equal sign, what you do to one side, you do to the other.
First, subtract 9x & 6 from both sides.
12x (-9x) + 6 (-6) > 9x (-9x) + 12 (-6)
12x - 9x > 12 - 6
Simplify.
12x - 9x > 12 - 6
3x > 6
Isolate the variable (x). Divide 3 from both sides.
(3x)/3 > (6)/3
x > 6/3
x > 2
x > 2 is your answer.
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Answer:
h=13z-30+ 27/z
Step-by-step explanation:
the can is 5 inches tall, so its height is 5 inches, so 3 inches off 5 inches that'd be 3/5.
the can has a diameter of 0.8 inches, meaning it has a radius of half that, or 0.4 inches.
if say the volume of iced-tea is V, how much is 3/5 of V? well, is just their product.
![\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=0.4\\ h=5 \end{cases}\implies V=\pi (0.4)^2(5)\implies V=0.8\pi \\\\\\ \stackrel{\textit{and 3/5 of that will be}}{\cfrac{3}{5}\cdot 0.8\pi }\implies V\approx \stackrel{in^3}{1.51}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cylinder%7D%5C%5C%5C%5C%20V%3D%5Cpi%20r%5E2%20h~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D0.4%5C%5C%20h%3D5%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Cpi%20%280.4%29%5E2%285%29%5Cimplies%20V%3D0.8%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Band%203%2F5%20of%20that%20will%20be%7D%7D%7B%5Ccfrac%7B3%7D%7B5%7D%5Ccdot%200.8%5Cpi%20%7D%5Cimplies%20V%5Capprox%20%5Cstackrel%7Bin%5E3%7D%7B1.51%7D)
The vertices A(-2,-1), B(-3, 2), C(-1, 3), and D(0, 0) form a parallelogram. The vertices A'(-1, -2), B'(2, -3), C'(3,-1),
attashe74 [19]
Answer:
A is correct.
Step-by-step explanation:
The rule for this transformation is that (x,y) will become (y,x) if it flips over y=x OR the y-axis and the x-axis.
Answer:
A) Radius: 3.44 cm.
Height: 6.88 cm.
B) Radius: 2.73 cm.
Height: 10.92 cm.
Step-by-step explanation:
We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.
a) We can express the volume of the soda can as:

This is the constraint.
The function we want to minimize is the surface, and it can be expressed as:

To solve this, we can express h in function of r:

And replace it in the surface equation

To optimize the function, we derive and equal to zero
![\frac{dS}{dr}=512*(-1)*r^{-2}+4\pi r=0\\\\\frac{-512}{r^2}+4\pi r=0\\\\r^3=\frac{512}{4\pi} \\\\r=\sqrt[3]{\frac{512}{4\pi} } =\sqrt[3]{40.74 }=3.44](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdr%7D%3D512%2A%28-1%29%2Ar%5E%7B-2%7D%2B4%5Cpi%20r%3D0%5C%5C%5C%5C%5Cfrac%7B-512%7D%7Br%5E2%7D%2B4%5Cpi%20r%3D0%5C%5C%5C%5Cr%5E3%3D%5Cfrac%7B512%7D%7B4%5Cpi%7D%20%5C%5C%5C%5Cr%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B512%7D%7B4%5Cpi%7D%20%7D%20%3D%5Csqrt%5B3%5D%7B40.74%20%7D%3D3.44)
The radius that minimizes the surface is r=3.44 cm.
The height is then

The height that minimizes the surface is h=6.88 cm.
b) The new equation for the real surface is:

We derive and equal to zero
![\frac{dS}{dr}=512*(-1)*r^{-2}+8\pi r=0\\\\\frac{-512}{r^2}+8\pi r=0\\\\r^3=\frac{512}{8\pi} \\\\r=\sqrt[3]{\frac{512}{8\pi}}=\sqrt[3]{20.37}=2.73](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdr%7D%3D512%2A%28-1%29%2Ar%5E%7B-2%7D%2B8%5Cpi%20r%3D0%5C%5C%5C%5C%5Cfrac%7B-512%7D%7Br%5E2%7D%2B8%5Cpi%20r%3D0%5C%5C%5C%5Cr%5E3%3D%5Cfrac%7B512%7D%7B8%5Cpi%7D%20%5C%5C%5C%5Cr%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B512%7D%7B8%5Cpi%7D%7D%3D%5Csqrt%5B3%5D%7B20.37%7D%3D2.73)
The radius that minimizes the real surface is r=2.73 cm.
The height is then

The height that minimizes the real surface is h=10.92 cm.