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maxonik [38]
3 years ago
8

A rancher purchased an SUV for $33,714 and made a down payment of 15% of the cost. The balance was financed for 4 years at an an

nual interest rate of 7%. Find the monthly truck payment. ​
Mathematics
1 answer:
seropon [69]3 years ago
3 0

Formula for monthly payment is:

A = P x (r(1+r)^t)/((1+r)^t-1) where P is the amount financed, r is the interest rate divided by 12 and t is the amount of time for the loan in months.

P = 33714 x 0.85 = 28656.90

A = 28656.90 x (0.07/12 (1+0.07/12)^48) / (1 +0.07/12)^48 - 1)

A = $686.23

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At which root does the graph of f(x) = (x+4)6(x + 7)5 cross the x axis?
Licemer1 [7]

Answer:

x = -7

Step-by-step explanation:

Our function is f(x) =(x+4)^6(x+7)^5. Notice that our possible roots are when x + 4 = 0 and when x + 7 = 0. So, our roots are -4 and -7.

However, the power above x + 4 is even, meaning the graph will simply <em>touch</em> the x-axis at x = -4, but not pass through. The power above x + 7, though, is odd, which means the graph will <em>cross</em> the x-axis.

Thus, the answer is x = -7.

5 0
3 years ago
Read 2 more answers
Ms. Wilson invested ​$34000 in two​ accounts, one yielding 8​% interest and the other yielding 9​%. If she received a total of​
Digiron [165]

Answer:

Amount invested at 8 % rate = x = $ 15000

Amount invested at 9 % rate = 34000 - x = 34000 - 15000 = $ 19000

Step-by-step explanation:

Total Amount = $ 34000

Let amount invested at 8 % rate = x

Amount invested at 9 % rate = $ 34000 - x

Total interest = $ 2910

2910 = \frac{x (8) (1)}{100} + \frac{(34000-x) (9) (1)}{100}

291000 = 8 x + 306000 - 9 x

x = 306000 - 291000

x = 15000

So  amount invested at 8 % rate = x = $ 15000

Amount invested at 9 % rate = 34000 - x = 34000 - 15000 = $ 19000

3 0
3 years ago
Which of these is not a method used to calculate finance charges?
Nikolay [14]

Answer:

previous balance

Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
2 years ago
Who can help and explain these 2 questions
Dominik [7]
Coefficients are added together because they are like terms, this can be proven with the distributive property. For example, x(2x+x)=2x^2+x^2=3x^2.

The commutative property of addition and the associative property demonstrate this.

The word "commutative" comes from "commute" or "move around", so the Commutative Property<span> is the one that refers to moving values around.
</span>
The associative property<span> states that you can add or multiply regardless of how the numbers are grouped. </span>
6 0
4 years ago
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