Check the picture below.
so, bearing in mind that, the radius and height are two sides in a right-triangle, thus both are at a ratio of each other, thus the radius is at 3:4 ratio in relation to the height.
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}\quad \begin{cases} r=3\\ h=4 \end{cases}\implies \stackrel{full}{V}=\cfrac{\pi \cdot 3^2\cdot 4}{3}\implies \stackrel{full}{V}=12\pi \\\\\\ \stackrel{\frac{2}{3}~full}{V}=12\pi \cdot \cfrac{2}{3}\implies \stackrel{\frac{2}{3}~full}{V}=8\pi ~in^3 \\\\\\ \textit{is draining water at a rate of }12~\frac{in^3}{min}\qquad \cfrac{8\pi ~in^3}{12~\frac{in^3}{min}}\implies \cfrac{2\pi }{3}~min](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D%5Cquad%20%0A%5Cbegin%7Bcases%7D%0Ar%3D3%5C%5C%0Ah%3D4%0A%5Cend%7Bcases%7D%5Cimplies%20%5Cstackrel%7Bfull%7D%7BV%7D%3D%5Ccfrac%7B%5Cpi%20%5Ccdot%203%5E2%5Ccdot%204%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bfull%7D%7BV%7D%3D12%5Cpi%20%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7B%5Cfrac%7B2%7D%7B3%7D~full%7D%7BV%7D%3D12%5Cpi%20%5Ccdot%20%5Ccfrac%7B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Cfrac%7B2%7D%7B3%7D~full%7D%7BV%7D%3D8%5Cpi%20~in%5E3%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bis%20draining%20water%20at%20a%20rate%20of%20%7D12~%5Cfrac%7Bin%5E3%7D%7Bmin%7D%5Cqquad%20%5Ccfrac%7B8%5Cpi%20~in%5E3%7D%7B12~%5Cfrac%7Bin%5E3%7D%7Bmin%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Cpi%20%7D%7B3%7D~min)
![\bf \textit{it takes }\frac{2\pi }{3}\textit{ minutes to drain }\frac{2}{3}\textit{ of it, leaving only }\frac{1}{3}\textit{ in it}\\\\ -------------------------------\\\\ \stackrel{\frac{1}{3}~full}{V}=12\pi \cdot \cfrac{1}{3}\implies \stackrel{\frac{1}{3}~full}{V}=4\pi\impliedby \textit{what's \underline{h} at this time?} \\\\\\ V=\cfrac{\pi r^2 h}{3}\quad \begin{cases} r=\frac{3h}{4}\\ V=4\pi \end{cases} \implies 4\pi =\cfrac{\pi \left( \frac{3h}{4} \right)^2 h}{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bit%20takes%20%7D%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%5Ctextit%7B%20minutes%20to%20drain%20%7D%5Cfrac%7B2%7D%7B3%7D%5Ctextit%7B%20of%20it%2C%20leaving%20only%20%7D%5Cfrac%7B1%7D%7B3%7D%5Ctextit%7B%20in%20it%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cstackrel%7B%5Cfrac%7B1%7D%7B3%7D~full%7D%7BV%7D%3D12%5Cpi%20%5Ccdot%20%5Ccfrac%7B1%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Cfrac%7B1%7D%7B3%7D~full%7D%7BV%7D%3D4%5Cpi%5Cimpliedby%20%5Ctextit%7Bwhat%27s%20%5Cunderline%7Bh%7D%20at%20this%20time%3F%7D%0A%5C%5C%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D%5Cquad%20%0A%5Cbegin%7Bcases%7D%0Ar%3D%5Cfrac%7B3h%7D%7B4%7D%5C%5C%0AV%3D4%5Cpi%20%0A%5Cend%7Bcases%7D%20%5Cimplies%204%5Cpi%20%3D%5Ccfrac%7B%5Cpi%20%5Cleft%28%20%5Cfrac%7B3h%7D%7B4%7D%20%5Cright%29%5E2%20h%7D%7B3%7D)
First get them all on one side
add 7x+60 to both sides
0=x^2+17x+60
now factor
what 2 numbers mutltiply to get 60 and add to get 17?
5 and 12
0=(x+5)(x+12)
set each to zero
0=x+5
-5=x
0=x+12
-12=x
the factored form is (x+5)(x+12)=0
the solutions are x=-12 and -5
He rode 9/10 mile
2/5 mile = 4/10 mi
1/2 mi = 5/10 mi
4/10 + 5/10 = 9/10 mi
Answer:
(3, 1)
Step-by-step explanation:
Answer:
1,3,5,6
Step-by-step explanation: