Answer:
picture is blocked for me because i'm using a school computer sorry bro
Step-by-step explanation:
Secure Unsecured
Credit APR APR
FAIR 7.00 7.65
Brand new Used Car
Purchase Price: 19,072 15,635
Sales Tax 4.5%
(price x 4.5%) <u> 858.24 </u> <u> 703.58</u>
Total price 19,930.24 16,338.58
Downpayment <u> (1,200.00) </u> <u> (1,200.00)</u>
For financing 18,730.24 15,138.58
Interest: Brand New:
Secured : 18,730.24 * 7% * 1/12 = 109.26
Unsecured: 18,730.24 * 7.65% * 1/12 = 119.41
Interest: Used Car
Secured: 15,138.58 * 7% * 1/12 = 88.31
Unsecured: 15,138.58 * 7.65% * 1/12 = 96.51
Answer:
with that i solved the answer is 5 days. what could be wrong
1.vertex form is 
to get into vertex form, complete the square
step 1: group x terms
step 2: factor coefient in front of x^2 term
prep for next step: take 1/2 of linear (1st dgree term) and square it
, 
step 3: add positive and negative of that previous number inside the parenthasees
factor perfect squaer trionmial
take out that -36 by expansion
2. factor out the 2
what 2 numbers multiply to get 12 and add to get 8?
2 and 6
3.
for 
in our case
a=1, b=-7, c=-6
so 
and 
4. remember that 
so 
treat 
as a variable
[tex}16[/tex]
Answer:
The integrals was calculated.
Step-by-step explanation:
We calculate integrals, and we get:
1) ∫ x^4 ln(x) dx=\frac{x^5 · ln(x)}{5} - \frac{x^5}{25}
2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}
3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}
4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}
5) \int\limits^{π/8}_0 {cos^4 (2x) } \, dx =\frac{sin(8x} + 8sin(4x)+24x}{6}=
=\frac{3π+8}{64}
6) ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x
7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}
8) ∫ tan^5 (x) sec(x) dx = \frac{sec^5 (x)}{5} -\frac{2sec^3 (x)}{3}+ sec x
