Sorry but you are cheating yourself but cheating in exams
I know but I will not say you real understand by yourself cheating is big than dying
Let the equation be:
y = ax^2 + bx + c.
Then, substitue the three points into the equation.
First point: 0 = a0^2 + b0 + c.
So c = 0.
Second point: -2 = a(-1)^2 + b(-1) + c.
So a - b + c = -2.
Third point: 6 = a*1^2 + b*1 + c.
So a + b + c = 6.
We know that c=0 already, so we substitute c=0 into the last two equations and we would get:
a - b = -2
a + b = 6
We add the two equations and we get:
2a = 4
a = 2
Then, we substitute a=2 into a-b=-2 and we get:
-b = -4
b = 4
Now we know a = 2, b = 4, and c = 0
Then, the equation of the parabola would be:
2x^2 + 4x
<h2>
Explanation:</h2>
Here we have to solve a system of linear equations in three variables. Let:
x: Cost of pens
y: Cost of pencils
We know that:
- Bailey buys 2 pens and 6 pencils and spends a total of $7.00. So:
![2x+6y=7 \ ... \ eq1](https://tex.z-dn.net/?f=2x%2B6y%3D7%20%5C%20...%20%5C%20eq1)
- Jeremiah buys 3 pens and 5 pencils and spends a total of $7.50. So:
![3x+5y=7.5 \ ... \ eq2](https://tex.z-dn.net/?f=3x%2B5y%3D7.5%20%5C%20...%20%5C%20eq2)
Multiplying eq1 by 3 and eq 2 by -2:
![\bullet \ 3(2x+6y)=3(7) \\ \\ 6x+18y=21 \\ \\ \\ \bullet \ -2(3x+5y)=-2(7.5) \\ \\ -6x-10=-15](https://tex.z-dn.net/?f=%5Cbullet%20%5C%203%282x%2B6y%29%3D3%287%29%20%5C%5C%20%5C%5C%206x%2B18y%3D21%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cbullet%20%5C%20-2%283x%2B5y%29%3D-2%287.5%29%20%5C%5C%20%5C%5C%20-6x-10%3D-15)
Then adding these two resultant equations:
![6x+18y=21 \\ \\ -6x-10y=-15 \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ \\ 8y=6 \\ \\ y=0.75 \\ \\ \\ So \ from \ eq \ 1: \\ \\ 2x+6(0.75)=7 \\ \\ 2x=7-4.5 \\ \\ 2x=2.5 \\ \\ x=1.25](https://tex.z-dn.net/?f=6x%2B18y%3D21%20%5C%5C%20%5C%5C%20-6x-10y%3D-15%20%5C%5C%20%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%5C_%20%5C%5C%20%5C%5C%208y%3D6%20%5C%5C%20%5C%5C%20y%3D0.75%20%5C%5C%20%5C%5C%20%5C%5C%20So%20%5C%20from%20%5C%20eq%20%5C%201%3A%20%5C%5C%20%5C%5C%202x%2B6%280.75%29%3D7%20%5C%5C%20%5C%5C%202x%3D7-4.5%20%5C%5C%20%5C%5C%202x%3D2.5%20%5C%5C%20%5C%5C%20x%3D1.25)
How much will Maggie spend to purchase 4 pens?
Each pen costs $1.24, so 4 pens will cost:
![\text{Cost of 4 pens}=4(1.25)=\boxed{\$5}](https://tex.z-dn.net/?f=%5Ctext%7BCost%20of%204%20pens%7D%3D4%281.25%29%3D%5Cboxed%7B%5C%245%7D)
The equation for compound interest is:
![A = P(1+ \frac{r}{n})^{ nt}](https://tex.z-dn.net/?f=A%20%3D%20P%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%29%5E%7B%20nt%7D%20%20)
Where r is the interest rate and n is the number of times per year it's applied.
Annually n = 1 and 7% interest r = 0.07
The quarterly rate 2% is already quartered 0.02 = r/n .
![(1+0.07)= (1+0.02 ) ^{4} \\ \\ 1.07 = (1.02) ^{4} \\ \\ 1.07 \neq 1.082](https://tex.z-dn.net/?f=%281%2B0.07%29%3D%20%281%2B0.02%20%29%20%5E%7B4%7D%20%20%5C%5C%20%20%5C%5C%201.07%20%3D%20%281.02%29%20%5E%7B4%7D%20%20%20%5C%5C%20%20%5C%5C%20%0A1.07%20%20%5Cneq%201.082%20)
You can see that Alexander is incorrect. A quarterly compound interest rate of 2% will accrue more interest than a 7% compound annual interest rate.
![ (1+0.07) = (1+ r) ^{4} \\ \\ 1.07 = (1+r) ^{4} \\ \\ \sqrt[4]{1.07} = r \\ \\ \sqrt[4]{1.07} - 1 = r \\ \\ r = 0.017](https://tex.z-dn.net/?f=%0A%281%2B0.07%29%20%3D%20%281%2B%20r%29%20%5E%7B4%7D%20%5C%5C%20%5C%5C%201.07%20%3D%20%281%2Br%29%20%5E%7B4%7D%20%5C%5C%20%5C%5C%20%5Csqrt%5B4%5D%7B1.07%7D%20%3D%20r%20%5C%5C%20%5C%5C%20%5Csqrt%5B4%5D%7B1.07%7D%20-%201%20%3D%20r%20%5C%5C%20%5C%5C%20r%20%3D%200.017%20)
1.7% compound quarterly
Answer:
D,C
Step-by-step explanation: