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2 years ago

Pleasee help meeee guysss help

Mathematics

1 answer:

Llana [10]2 years ago

6 0

Answer:

A is the true answer and others are wrong

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WILL GIVE BRAINLIEST Mr. and Mrs. Bailey hope to send their son to college in eleven years. How much money should they invest no

Ronch [10]

Mr. and Mrs. Bailey need to invest $2906.50 so as to send their son to college.

<h3>Compound interest</h3>

Compound interest is given by:

$A=P(1+\frac{r}{n})^{nt}$

where A is the amount after t years, P is initial amount, r is the rate and n is the times compounded per period

Given that n = 1, r = 9% = 0.09, A = $7500 t = 11. Hence:

$7500=P(1+\frac{0.09}{1} )^{1*11}\\\\P=\$2906.50$

Mr. and Mrs. Bailey need to invest $2906.50 so as to send their son to college.

Find out more on Compound interest at: brainly.com/question/24924853

7 0

3 years ago

Plzzzzz help I need

Lera25 [3.4K]

Answer:

i dont know tbh

Step-by-step explanation:

good luck

8 0

3 years ago

Plot four lines. Each line should pass through the point

notka56 [123]

Answer:

Step-by-step explanation:

I am on the same question

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

How many lengths of pipe 2 1/3 ft long can be cut from a pipe that is 63 ft long? Assume there is no kerf.

Zinaida [17]

The answer is 27ft long

7 0

3 years ago