One approach that might be a great help would be if you would look
at the choices ... those things at the bottom of the question that you
didn't give us.
-- One leg is 3 times the length of the other leg.
-- Any right triangle that has one leg three times as long as the other
leg is similar to this triangle.
I'll bet that if you look through the choices, you'll find one there.
To effectively determine the correct answer, it would be helpful to write this into an algebraic expression. We let x as the number. We do as follows:
<span>Four times the square of a certain number increased by 6 times the number equals 108.
4x^2 + 6x = 108
The numbers can be either of the following since the equation generated was a quadratic equation which has two roots.
x = 4.5
x = -6 </span>
Answer:
Step-by-step explanation:
Check the picture below.
so we know the radius of the semicircle is 2 and the rectangle below it is really a 4x4 square, so let's just get their separate areas and add them up.
![\stackrel{\textit{area of the semicircle}}{\cfrac{1}{2}\pi r^2}\implies \cfrac{1}{2}(\stackrel{\pi }{3.14})(2)^2\implies 3.14\cdot 2\implies 6.28 \\\\\\ \stackrel{\textit{area of the square}}{(4)(4)}\implies 16 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{sum of both areas}}{16+6.28=22.28}~\hfill](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20semicircle%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28%5Cstackrel%7B%5Cpi%20%7D%7B3.14%7D%29%282%29%5E2%5Cimplies%203.14%5Ccdot%202%5Cimplies%206.28%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20square%7D%7D%7B%284%29%284%29%7D%5Cimplies%2016%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20both%20areas%7D%7D%7B16%2B6.28%3D22.28%7D~%5Chfill)
In the old aquarium, 3 tablespoons of salt were used for 20 gallons. That's a conversion ratio of
That is, for every gallon of water in the aquarium, there was 0.15 tbsp of salt. So a 50 gallon aquarium, Ill should add
to maintain the salinity.
