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3 years ago

Help I don't understand

Mathematics

1 answer:

Leona [35]3 years ago

8 0

150 student voted for later start.

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A savings account earns 5% simple interest per year. The principal is $1200. What is the balance after 4 years?

stiv31 [10]

The formula of the Simple Interest is:
I=PRT
P for Principle Amount ($1200)
R for Rate (5%= $\frac{5}{100}$ = 0.05)
T for Time in years (4 years)
I = 1200 × 0.05 × 4
= $240

Add the interest to the principle amount to check the balance
$1200 + $240 = $1440

7 0

4 years ago

Read 2 more answers

A Web music store offers two versions of a popular song. The size of the standard version is 2.1 megabytes (MB). The size of the

liq [111]

If the number of downloads of the standard version is x, and the high quality is x, 2.1*x+4.1*y=2761 (not 1010 due to that this is multiplied by 2.1 and 4.1, therefore representing the total amount of megabytes) In addition, there are 1010 total downloads, and it's either 2.1 MB or 4.1 MB, so x+y=1010.

We have
2.1x+4.1y=2761
x+y=1010

Multiplying the second equation by -2.1 and adding it to the first equation, we get 2y=2761-1010*2.1=640 and by dividing both sides by 2 we get y=320 downloads of the high quality version

5 0

3 years ago

2−6(−5t+1)

VashaNatasha [74]

Answer: I don't think any of those choices are correct.

Step-by-step explanation:

2-6*-5t+6

6*-5t+4

6 0

3 years ago

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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

Determine the solution to the system of equations

gizmo_the_mogwai [7]

Answer:

no sé pero no se y no se y más no se

3 0

3 years ago