First, you should solve for
, which equals
. Now, solve the integral of
=
, to get that
. You can check this by taking the integral of what you got. Now by the Fundamental Theorem
![\int\limits^2_0 {4x} \, dx=[2x^2] ^{2}_{0}=2(2)^{2}-2(0)^2=8](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_0%20%7B4x%7D%20%5C%2C%20dx%3D%5B2x%5E2%5D%20%5E%7B2%7D_%7B0%7D%3D2%282%29%5E%7B2%7D-2%280%29%5E2%3D8)
.
This should be the answer to your question, if I understood what you were asking correctly.
Answer: D
Step-by-step explanation:
Answer:
- x = 0 or 1
- x = ±i/4
- x = -5 (twice)
Step-by-step explanation:
Factoring is aided by having the equations in standard form. The first step in each case is to put the equations in that form. The zero product property tells you that a product is zero when a factor is zero. The solutions are the values of x that make the factors zero.
1. x^2 -x = 0
x(x -1) = 0 . . . . . x = 0 or 1
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2. 16x^2 +1 = 0
This is the "difference of squares" ...
(4x)^2 - (i)^2 = 0
(4x -i)(4x +i) = 0 . . . . . x = -i/4 or i/4 (zeros are complex)
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3. x^2 +10x +25 = 0
(x +5)(x +5) = 0 . . . . . x = -5 with multiplicity 2
So, r' (the new r) becomes kr, and h'=h*k,
so V' = pi * (r')^2*h' = pi * (r*k)^2*(h*k) = pi*r^2*h*k^3 = V * k^3
Convinced?
She’ll have to make 4 more stitches in order to get 6 good ones. Each time she creates 3 she undoes 2 which after 4 more times would leave her with 6 good stitches.
