Some racquet balls are sold in cylindrical cans of three balls. Each ball has a diameter of 2.25 inches. The can has a diameter
2 answers:
In order to find the empty space we need to find the total space and the filled space.
The formula we use to find the volume of the cans or any cylinder is:
Plug in the givens:
Simplify:
Now let's find the balls' volume. These cans can hold 3 balls:
Take the cube:
Solve:
To find the empty space:
The empty space is 8.94 in3.
The formula we use to find the volume of the cans or any cylinder is:
Plug in the givens:
Simplify:
Now let's find the balls' volume. These cans can hold 3 balls:
Take the cube:
Solve:
To find the empty space:
The empty space is 8.94 in3.
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Ooh, fun
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.
B.
sepearte the integrals
![\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-5%7D_%7B-4%7D%20%7Bx%5E2%2Bx-12%7D%20%5C%2C%20dx%20%3D%20%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-5%7D_%7B-4%7D%3D%28%5Cfrac%7B-125%7D%7B3%7D%2B%5Cfrac%7B25%7D%7B2%7D%2B60%29-%28%5Cfrac%7B64%7D%7B3%7D%2B8%2B48%29%3D%5Cfrac%7B23%7D%7B6%7D)
next one
![\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-4%7D_3%20%7B-x%5E2-x%2B12%7D%20%5C%2C%20dx%3D-1%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-4%7D_%7B3%7D%3D-1%28%28-64%2F3%29%2B8%2B48%29-%289%2B%289%2F2%29-36%29%29%3D%5Cfrac%7B343%7D%7B6%7D)
the last one you can do yourself, it is
the sum is
so the area under the curve is
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.
B.
sepearte the integrals
next one
the last one you can do yourself, it is
the sum is
so the area under the curve is
The sides of a right triangle are
1995. 164 million
2001. 169 million
t = 2001 - 1995 = 6
169 =
![p^{6} = \frac{169}{164}= 1.0304878 \\ p = \sqrt[6]{1.0304878}](https://tex.z-dn.net/?f=%20p%5E%7B6%7D%20%3D%20%5Cfrac%7B169%7D%7B164%7D%3D%201.0304878%20%5C%5C%20%20p%20%3D%20%20%5Csqrt%5B6%5D%7B1.0304878%7D%20%20)
p = 1.00502
t 1 = 2015 - 2001 = 14
f ( t1 ) =
f (t1) = 169 * 1.0762 = 181.27
Answer: a country´s population in 2015 will be 181 million.
2001. 169 million
t = 2001 - 1995 = 6
169 =
p = 1.00502
t 1 = 2015 - 2001 = 14
f ( t1 ) =
f (t1) = 169 * 1.0762 = 181.27
Answer: a country´s population in 2015 will be 181 million.