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3 years ago

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You start with $1000 and every day you will receive the previous day’s total plus an additional $1000 a day for 30 days. How muc

h money will you have after 30 days?

1 answer:

Roman55 [17]3 years ago

4 0

Answer:

The total earned after 30 days is 465000.

Step-by-step explanation:

The amount of money that I'll recieve can be modelled as a arithmetic sequence in which the next element is related to the prior by a sum of a rate "q". This sequence can be seen below:

{1000, 2000, 3000, ...}

Where q = 1000. In order sum all the "n" terms on a sequence of this kind we can use the following formula:

Sn = (n/2)*(a1 + an)

And to find the term an, we can use the formula:

an = a0 + (n-1)*r

Where n is the position of the number we want to calculate, in this case 30, a0 is the first number on the sequence and r is the rate between consecutive numbers. So the 30th term is:

a30 = 1000 + (30 -1)*1000

a30 = 1000 + 29*1000

a30 = 1000 + 29000 = 30000

And the total obtained is:

s30 = (30/2)*(1000 + 30000) = (30/2)*(31000)

s30 = 930000/2 = 465000

The total earned after 30 days is 465000.

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Find the area of a segment of a circle if its arc is 120 degrees and the radius is 3 inches.

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Answer:<u> 9.42 inches</u>, 3π in terms of pi

Step-by-step explanation:

A full circle is 360 degrees, so if the segment has an arc of 120 degrees, it is 1/3 of a regular circle.

To find the area of a circle, we use the equation A = πr^2. In this case, we will divide the equation by 3 because we are only finding 1/ of a circle. So, our equation will be A = (πr^2)/3.

Using 3.14 for π:

A = (πr^2)/3.

A = (π3^2)/3

A = (π9)/3

A = 3π

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2 years ago

a counter top is in the shape of a trapezoid. The lengths of the bases are 70 1/2 and 65 1/2 inches long. The area of the counte

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For the trapezoid, the equation used to solve for the area is,
A = (0.5)(b₁ + b₂)(h)
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3 years ago

A shelf has 20 bags of potatoes. Two of the bags have potatoes starting to go bad. The probability of drawing a bag with bad pot

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Probability of drawing bad potatoes = 0.1

The probability of drawing a bag with bad potatoes is less than 1.

The probability of drawing a bag with good potatoes is greater than 0

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probability of bad potatoes = 2/20

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3 years ago

Read 2 more answers

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

3 years ago