A varies inversely as the square of b. If a is 1 when b is 4, find a when b is 4.<br> with solution

ss7ja [257]

Answer:

hshshxhdjebrbhdjwjdbevvdhdhfiiehrvrvdhdjrirubrvfgdhjdjejrhvrvdghfueyrggdvdvdgdhdudhdhegdghdhhrhdbdbdbhdhdhfhfydhhdbfbdbdbrbrhhrhrhrhfhfuyfh4h4bbdbdhxufhfbbdhdhdhrhffbfbdbduc

7 0

3 years ago

ITS AN EMERGENCY PLEASE!

Lady bird [3.3K]

I’m pretty sure the power is 4

3 0

3 years ago

Otto invests $1,000 at age 20. Otto wants theinvestment to be worth $4,000 by age 30. Ifinterest is compounded continuously, wha

AveGali [126]

$r\approx0.139$

1) Since this is a Continuously Compounded operation in a 10 yrs period, then we can write out the following equation:

$\begin{gathered} A=Pe^{rt} \\ \end{gathered}$

2) Plugging into the equation the given data and since Otto is 20 yrs old and he plans to get $4,000 in ten years, we can write out:

$\begin{gathered} 4000=1000e^{\mleft\{10r\mright\}} \\ \frac{4000}{1000}=\frac{1000e^{10r}}{1000} \\ 4=e^{10r} \\ \ln (4)=\ln (e)^{10r} \\ 10r=\ln (4) \\ r=\frac{\ln (4)}{10} \\ r=0.1386 \end{gathered}$

3) Thus the rate Otto needs is

$r=0.1386\approx0.139$

Note that since the 0.1386 the six here is greater than 5 then we can round up to the next thousandth, in this case: 0.139. For the 0.1386 is closer to 0.139 (0.004) than to 0.138 (0.006).

Or 13.9%

8 0

1 year ago

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

N is a positive intiger. Explain why 2n+1 must be an odd number.

pogonyaev

Answer:

Because the number would be multiplied by an even number, which is 2. If the number (odd or even) is multiplied by two, then the result would be an <em>even number</em>.

Proof:

2(1) = 2

2(6) = 12

2(7) = 14

2(3) = 6

The number would then be added by an odd number, which is 1. If an even number is added with an odd number, then the result would be an odd number.

Proof:

2 + 1 = 3

12 + 1 = 13

14 + 1 = 15

6 + 1 = 7

6 0

3 years ago

A varies inversely as the square of b. If a is 1 when b is 4, find a when b is 4. with solution