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3 years ago

14

in intjammera received a $3000 car loan she plans on paying off the loan in 2 years at the end of 2 years jamerra will have paid

$450 in interest what is the simple interest rate on the car loan?

1 answer:

mel-nik [20]3 years ago

4 0

Answer:

450=(3000*r*2)/100

solve for r

therefore r = 7.5 per cent

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fourteen boys and 21 girls will be equally divided into groups. Find the greatest number of groups that can be created if no one

shtirl [24]

Well if you add 14 and 21 you get 35. with 35 you can have 7 groups of 5 but if they have to be equal with the number of boys and girls in each group it would still be 7. two boys in each group with three girls in each group.

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4 years ago

What is the length of the side of a right triangle that has a side length of 12in and a hypotenuse that measures 15ft

Viktor [21]

Answer: 14.96ft

Step-by-step explanation:

1. To solve this exercise you must apply the Pythagorean Theorem, which is:

$a=\sqrt{b^{2}+c^{2}}$

Where $a$ is the hypotenuse, and $b$ and $c$ are the other sides of the triangle.

2. You have that the sides given measures 12 inches. 1 feet has 12 inches. Therefore, this side measures 1 feet.

2. Then, when you solve for one of the sides and substitute the values given in the problem into the formula shown above, you obtain that the length of the side of the rigth triangle is:

$b=\sqrt{(15ft)^{2}-(1ft)^{2}}$

$b=4\sqrt{14}=14.96ft$

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Number 11 and 14, I need help on please.

Ulleksa [173]

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3 years ago

If you are given a3=2 a5=16, find a100.

ra1l [238]

I suppose $a_n$ denotes the n-th term of some sequence, and we're given the 3rd and 5th terms $a_3=2$ and $a_5=16$ . On this information alone, it's impossible to determine the 100th term $a_{100}$ because there are infinitely many sequences where 2 and 16 are the 3rd and 5th terms.

To get around that, I'll offer two plausible solutions based on different assumptions. So bear in mind that this is not a complete answer, and indeed may not even be applicable.

• Assumption 1: the sequence is arithmetic (a.k.a. linear)

In this case, consecutive terms <u>d</u>iffer by a constant d, or

$a_n = a_{n-1} + d$

By this relation,

$a_{n-1} = a_{n-2} + d$

and by substitution,

$a_n = (a_{n-2} + d) + d = a_{n-2} + 2d$

We can continue in this fashion to get

$a_n = a_{n-3} + 3d$

$a_n = a_{n-4} + 4d$

and so on, down to writing the n-th term in terms of the first as

$a_n = a_1 + (n-1)d$

Now, with the given known values, we have

$a_3 = a_1 + 2d = 2$

$a_5 = a_1 + 4d = 16$

Eliminate $a_1$ to solve for d :

$(a_1 + 4d) - (a_1 + 2d) = 16 - 2 \implies 2d = 14 \implies d = 7$

Find the first term $a_1$ :

$a_1 + 2\times7 = 2 \implies a_1 = 2 - 14 = -12$

Then the 100th term in the sequence is

$a_{100} = a_1 + 99d = -12 + 99\times7 = \boxed{681}$

• Assumption 2: the sequence is geometric

In this case, the <u>r</u>atio of consecutive terms is a constant r such that

$a_n = r a_{n-1}$

We can solve for $a_n$ in terms of $a_1$ like we did in the arithmetic case.

$a_{n-1} = ra_{n-2} \implies a_n = r\left(ra_{n-2}\right) = r^2 a_{n-2}$

and so on down to

$a_n = r^{n-1} a_1$

Now,

$a_3 = r^2 a_1 = 2$

$a_5 = r^4 a_1 = 16$

Eliminate $a_1$ and solve for r by dividing

$\dfrac{a_5}{a_3} = \dfrac{r^4a_1}{r^2a_1} = \dfrac{16}2 \implies r^2 = 8 \implies r = 2\sqrt2$

Solve for $a_1$ :

$r^2 a_1 = 8a_1 = 2 \implies a_1 = \dfrac14$

Then the 100th term is

$a_{100} = \dfrac{(2\sqrt2)^{99}}4 = \boxed{\dfrac{\sqrt{8^{99}}}4}$

The arithmetic case seems more likely since the final answer is a simple integer, but that's just my opinion...

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2 years ago