Let i = sqrt(-1) which is the conventional notation to set up an imaginary number
The idea is to break up the radicand, aka stuff under the square root, to simplify
sqrt(-8) = sqrt(-1*4*2)
sqrt(-8) = sqrt(-1)*sqrt(4)*sqrt(2)
sqrt(-8) = i*2*sqrt(2)
sqrt(-8) = 2i*sqrt(2)
<h3>Answer is choice A</h3>
Answer:
The answer is
.
Step-by-step explanation:
The formula is supposed to be
or
, and if you do the math, the slope should be easy to solve.
Answer:
$56,226
Step-by-step explanation:
x = investment
Earning interest of 3.57% is the equivalent of multiplying the initial balance (investment) by 1.0357;
Over 20 years, the new balance can be found by multiplying the initial balance by (1.0357)²⁰;
We can formulate an equation to solve to get the initial investment:
x(1.0357)²⁰ = 113400
(2.01687752)x = 113400
x = ¹¹³⁴⁰⁰/₍₂.₀₁₆...₎
x = 56225.5263 → 56226
Answer:
Step-by-step explanation:
y = mx + b
slope(m) = 1/4
(-1,-4)...x = -1 and y = -4
now sub and find b, the y intercept
-4 = 1/4(-1) + b
-4 = -1/4 + b
-4 + 1/4 = b
-16/4 + 1/4 = b
-15/4 = b
so the equation in slope intercept form is : y = 1/4x - 15/4
but we need it in general form : Ax + By + C = 0
y = 1/4x - 15/4.....multiply everything by 4 to get rid of fractions
4y = x - 15......subtract x and add 15 ...to both sides
4y - x + 15 = 0 <=== general form
Given a solution

, we can attempt to find a solution of the form

. We have derivatives



Substituting into the ODE, we get


Setting

, we end up with the linear ODE

Multiplying both sides by

, we have

and noting that
![\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bx%28%5Cln%20x%29%5E2%5Cright%5D%3D%28%5Cln%20x%29%5E2%2B%5Cdfrac%7B2x%5Cln%20x%7Dx%3D%28%5Cln%20x%29%5E2%2B2%5Cln%20x)
we can write the ODE as
![\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bwx%28%5Cln%20x%29%5E2%5Cright%5D%3D0)
Integrating both sides with respect to

, we get


Now solve for

:


So you have

and given that

, the second term in

is already taken into account in the solution set, which means that

, i.e. any constant solution is in the solution set.