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4 years ago

13

Which scatterplots have a nonlinear association? Check all that apply. On a graph, points form a curved line. On a graph, points

form a curved line. On a graph, points are grouped together to form a vertical cluster. On a graph, points form a curved line. On a graph, points are scattered all over the graph.

2 answers:

Katen [24]4 years ago

8 0

one, two, and four.

i hAve tO wRiTe aTleAsT 2O wOrDs t0 sUbMit mY aNsWeR.

daser333 [38]4 years ago

3 0

Answer:

<em>Its B</em>

Step-by-step explanation:

Just took the test

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What's the value of given expression<br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B1%20%5Ctimes%201%7D%20" id="TexFormul

arsen [322]

Answer:

1

Step-by-step explanation:

1st step : multiple the number inside square root

1 x 1 =1

so then think about which time 1 will give you 1

that will 1

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3 years ago

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3 years ago

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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

inna [77]

Answer:

The value of 1 in Maishas time is hundredth

The value of the 1 in Pattis time tenth

Step-by-step explanation:

The place value of decimals begins at tenth and extends towards the right. Hence, the one decimal place in Pattis time has the value of tenth while the 1 in Maishas time occurring at the second decimal place has a place value of hundredth.

Hence the place value of decimals may be tenth, hundredth, thousandth etc depending on the number of decimal places and the digit of interest.

5 0

3 years ago

Given that x=6 and y= -4<br> 3x + x over y - y^2

soldi70 [24.7K]

Step-by-step explanation:

x=6

y=-4

3x+x / y-y^2

3(6)+6/-4-(-4)^2

24/-20

= -1.2

3 0

2 years ago