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3 years ago

Please help meeeee ASAP

Mathematics

1 answer:

Serga [27]3 years ago

3 0

Measure of angle K is 146 degrees.

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X + 16 = 9z Help please

diamong [38]

Answer:x=9z + 16

z= x/9 + 16/9

Step-by-step explanation: dont know if i solve for x or z so i did both subtract 16 from both sides for x

Isolate the variable by dividing each side by factors that don't contain the variable. for z let me know if u need help

3 0

3 years ago

Read 2 more answers

Find the exact area of the surface obtained by rotating the curve about the x-axis. y = 1 + ex , 0 ≤ x ≤ 9

tekilochka [14]

The surface area is given by

$\displaystyle2\pi\int_0^9(1+e^x)\sqrt{1+e^{2x}}\,\mathrm dx$

since $y=1+e^x\implies y'=e^x$ . To compute the integral, first let

$u=e^x\implies x=\ln u$

so that $\mathrm dx=\frac{\mathrm du}u$ , and the integral becomes

$\displaystyle2\pi\int_1^{e^9}\frac{(1+u)\sqrt{1+u^2}}u\,\mathrm du$

$=\displaystyle2\pi\int_1^{e^9}\left(\frac{\sqrt{1+u^2}}u+\sqrt{1+u^2}\right)\,\mathrm du$

Next, let

$u=\tan t\implies t=\tan^{-1}u$

so that $\mathrm du=\sec^2t\,\mathrm dt$ . Then

$1+u^2=1+\tan^2t=\sec^2t\implies\sqrt{1+u^2}=\sec t$

so the integral becomes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec t}{\tan t}+\sec t\right)\sec^2t\,\mathrm dt$

$=\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec^3t}{\tan t}+\sec^3 t\right)\,\mathrm dt$

Rewrite the integrand with

$\dfrac{\sec^3t}{\tan t}=\dfrac{\sec t\tan t\sec^2t}{\sec^2t-1}$

so that integrating the first term boils down to

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\frac{\sec t\tan t\sec^2t}{\sec^2t-1}\,\mathrm dt=2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\frac{s^2}{s^2-1}\,\mathrm ds$

where we substitute $s=\sec t\implies\mathrm ds=\sec t\tan t\,\mathrm dt$ . Since

$\dfrac{s^2}{s^2-1}=1+\dfrac12\left(\dfrac1{s-1}-\dfrac1{s+1}\right)$

the first term in this integral contributes

$\displaystyle2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\left(1+\frac12\left(\frac1{s-1}-\frac1{s+1}\right)\right)\,\mathrm ds=2\pi\left(s+\frac12\ln\left|\frac{s-1}{s+1}\right|\right)\bigg|_{\sqrt2}^{\sqrt{1+e^{18}}}$

$=2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}$

The second term of the integral contributes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\sec^3t\,\mathrm dt$

The antiderivative of $\sec^3t$ is well-known (enough that I won't derive it here myself):

$\displaystyle\int\sec^3t\,\mathrm dt=\frac12\sec t\tan t+\frac12\ln|\sec t+\tan t|+C$

so this latter integral's contribution is

$\pi\left(\sec t\tan t+\ln|\sec t+\tan t|\right)\bigg|_{\pi/4}^{\tan^{-1}(e^9)}=\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

Then the surface area is

$2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}+\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

$=\boxed{\left((2+e^9)\sqrt{1+e^{18}}-\sqrt2+\ln\dfrac{(e^9+\sqrt{1+e^{18}})(\sqrt{1+e^{18}}-1)}{(1+\sqrt2)(1+\sqrt{1+e^{18}})}\right)\pi}$

4 0

4 years ago

How do I find the volume of a square polyhedron

lawyer [7]

First, find the volume. For a square polyhedron, the formula for the volume is the following:

Volume=length×width×height

Hope this helps.

5 0

3 years ago

Read 2 more answers

You invested a total of $60,000 in 2 funds earning 8.5% and 10% simple interest. During 1 year, the 2 funds earned a total of $5

maksim [4K]

The amount invested in the fund earning 8.5% is $40,000 whereas $20,000 was invested in the fund earning 10%

What is the return on each fund?

The return on each fund is determined as the amount invested multiplied by the interest rate on the fund

;Let X be the amount invested at 8.5%

Interest=8.5%*X=0.085X

The interest of the fund invested at 10%=($60,000-X)*10%

The interest of the fund invested at 10%=$6000-0.10X

Total interest=0.085X+6000-0.10X

Total interest earned during year 1=$5,400

5400=0.085X+6000-0.10X

5400=6000-0.015X

0.015X=6000-5400

0.015X=600

X=600/0.015

X=$40,000(amount invested at 8.5%)

amount invested at 10%=$60,000-$40,000

amount invested at 10%=$20,000

Find out more about simple interest on:https://brainly.in/question/37840965

#SPJ1

7 0

2 years ago

If a tank filled with water contains a block and the height of the water above point A within the block is 0.6 meter, what's the

Zielflug [23.3K]

Hi! The answer is D. 5.88kPa

5 0

4 years ago