You must do 1/2 times base times height
Answer: The goal was $500
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Work Shown:
x% = x/100
175% = 175/100 = 1.75
Let g be the goal, which is the amount of money the club wanted to raise
175% of goal = 175% of g = 1.75g
The expression 1.75g represents how much money was actually raised, which was $875. Set the two expressions equal to each other. Solve for g
1.75g = 875
g = 875/1.75 ....... divide both sides by 1.75
g = 500
The club's goal was to raise $500
Note how 75% of $500 is 0.75*500 = 375
When they raised 175% of the goal, this means they went 75% overboard and added on 375 additional dollars (on top of the 500 they wanted). So they got to 500+375 = 875 which lines up with the instructions. This helps verify the answer.
Or we can see that 1.75*g = 1.75*500 = 875 which helps confirm the answer as well.
Answer:
Step-by-step explanation:
Final concentration % = [grams of salt / mililiters of solution] * 100
grams of salt = grams of salt from solution 1 + grams of salt from solution 2
grams of salt from solution 1 = mililiters * %/100 = 80 mililiters * 0.25 g/mililiters = 20 g
grams of salt from solution 2 = mililiters *%/100 = x*0.10 g/mililiters = 0.1x
mililiters of final solution = mililiters from solution 1 + mililiters from solution 2
mililiters of final solution = 80 mililiters + x mililiters
=> Final concentration, % = [0,10x + 20 g] / [x + 80 mililiters] * 100
![\bf f(x)=x+3x^{\frac{2}{3}}\implies \cfrac{dy}{dx}=1+3\left(\frac{2}{3}x^{-\frac{1}{3}} \right)\implies \cfrac{dy}{dx}=1+\cfrac{2}{\sqrt[3]{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\sqrt[3]{x}+2\implies -2=\sqrt[3]{x} \\\\\\ (-2)^3=x\implies \boxed{-8=x}\\\\ -------------------------------\\\\ 0=\sqrt[3]{x}\implies \boxed{0=x}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3Dx%2B3x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B3%5Cleft%28%5Cfrac%7B2%7D%7B3%7Dx%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D%20%20%5Cright%29%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B%5Ccfrac%7B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Csqrt%5B3%5D%7Bx%7D%2B2%5Cimplies%20-2%3D%5Csqrt%5B3%5D%7Bx%7D%0A%5C%5C%5C%5C%5C%5C%0A%28-2%29%5E3%3Dx%5Cimplies%20%5Cboxed%7B-8%3Dx%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A0%3D%5Csqrt%5B3%5D%7Bx%7D%5Cimplies%20%5Cboxed%7B0%3Dx%7D)
now, f(0) = 0, and f(-8) is an imaginary value or no real value.
now, f(-10) will also give us an imaginary value
and f(1) = 4
so, doing a first-derivative test on 0, is imaginary to the left and positive on the right, and before and after 1, is positive as well, so f(x) is going up on those intervals.
however, f(0) is 0 and f(1) is higher up, so the absolute maximum will have to be f(1), and we can use f(0) as a minimum, and since it's the only one, the absolute minimum.
the other two, the endpoint of -10 and the critical point of -8, do not yield any values for f(x).
