8640 divided by40 long division

Natalie and Preston would like to accumulate $25,000 at the end of 3.5 years for a future down payment on a house in Prescott. H

Brrunno [24]

Answer:

$7.4 should be deposited at the end of each week

Step-by-step explanation:

In this question, we are asked to calculate the principal amount that should be deposited to get a house of $25,000 at the interest and time given in the question

To solve this problem, we use the compound interest formula.

Mathematically;

A = P(1+ r/n) ^ nt

From the question, we identify the following;

A is the amount that is targeted which is the principal plus the targeted interest = $25,000 according to the question

P is the amount invested at the beginning which we are looking for

r is the rate which is 7.2% = 7.2/100 = 0.072

n is the number of times we compound per year which is 12 times( compounding is per month)

t is the number of years which is 3.5 years

To get the principal, we can rewrite the equation to be;

A/(1+r/n)^nt = P

25,000/(1 + 0.072/12)^(12 * 3.5) = P

P = 25,000/18.54 = $1,348

But the question asks how much to be deposited at the end of each week

The number of weeks in 3.5 years is 3.5 * 52 =182 weeks

Thus, the amount deposited at the end of each week will be 1,348/182 = $7.4

3 0

3 years ago

Evaluate 8P4 how can I evaluate this problem

Leno4ka [110]

NPr = n! / (n-r)!

8P4 = 8! / (8-4)!

= 8! / 4!

= 1,680

5 0

3 years ago

Read 2 more answers

What is the volume of the prism? enter your awnser in the box as a mixed number in simplist form

liberstina [14]

Answer:

67 1/2 is a mixed number

Step-by-step explanation:

The volume of a prism, is the area of the cross section X the length:

In this case it is base × height × length:

6 × 4.5 × 2.5 = 67.5cm^3

= 67 1 / 2 as a mixed number

4 0

3 years ago

Write in normal form 2.8 x 10⁶ ?

Aliun [14]

It would be 2,800,000

3 0

2 years ago

Read 2 more answers

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

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Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

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Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

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Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

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Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

2 years ago