This is a quadratic equation with a general equation of ax^2 + bx + c.
The quadratic formula can help to get the roots of the equation. We know the highest degree of that equation is 2; so there will be also two roots.
The quadratic formula is
x = [-b ± √(b^2 - 4ac)] / 2a
With a = 1, b = 7, c = 2,
x = {-7 ± √[(7)^2 - 4(1)(2)]} / 2(1) = (-7 ± √41) / 2
So the two roots are
x1 = (-7 + √41) / 2 = -0.2984
x2 = (-7 - √41) / 2 = 0.2984
This is also another way of factorizing the equation
(x + 0.2984)(x + 0.2984) = x^2 + 7x + 2
divide by 2 on both sides.
3/2
x=1.5
Answer:
Step-by-step explanation:
Incomplete question.
Answer: 
Step-by-step explanation:
Given
Julien's parents want $64,000 at the end of 3 years
The rate of interest is 10% annually
Suppose they invested P amount initially
So, compound interest, compounded monthly is
![\Rightarrow A=P\left( 1+\frac{r}{n}\right)^{nt}\quad [\text{n=Numner of compounding periods}]\\\\\text{Insert the values}\\\Rightarrow 64,000=P\left( 1+\frac{0.10}{12}\right)^{12\times 3}\\\\\Rightarrow 64,000=P\left( 1+\frac{10}{12}\right)^{36}\\\\\Rightarrow P=\dfrac{64000}{1.3481}\\\\\Rightarrow P=\$\ 47,471.37\approx \$ 47,471](https://tex.z-dn.net/?f=%5CRightarrow%20A%3DP%5Cleft%28%201%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%5Cquad%20%5B%5Ctext%7Bn%3DNumner%20of%20compounding%20periods%7D%5D%5C%5C%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5CRightarrow%2064%2C000%3DP%5Cleft%28%201%2B%5Cfrac%7B0.10%7D%7B12%7D%5Cright%29%5E%7B12%5Ctimes%203%7D%5C%5C%5C%5C%5CRightarrow%2064%2C000%3DP%5Cleft%28%201%2B%5Cfrac%7B10%7D%7B12%7D%5Cright%29%5E%7B36%7D%5C%5C%5C%5C%5CRightarrow%20P%3D%5Cdfrac%7B64000%7D%7B1.3481%7D%5C%5C%5C%5C%5CRightarrow%20P%3D%5C%24%5C%2047%2C471.37%5Capprox%20%5C%24%2047%2C471)
Answer:
(0,4) is the y intercept for function A
(0,3) is the y intercept for function B
so function A has the greatest y intercept
Step-by-step explanation:
