Can you please help me. I think it is x=0 and x=3

Mathematics

1 answer:

stiks02 [169]3 years ago

6 0

When the graph is constant, you have to look for when it draws a straight line.
Look at the x axis, since it asks for the x value.

Your answer is correct! When x = 0, to x = 3, the function is constant.

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A class consists of 12 boys and 12 girls. The teacher picks five students to present their work to the rest of the class

Varvara68 [4.7K]

Answer:

0.01863, yes preference

Step-by-step explanation:

given that a class consists of 12 boys and 12 girls. The teacher picks five students to present their work to the rest of the class and says that the five students are being selected at random

Here selecting any 5 students from the group of total 24 students is combination because order does not matter.

Total no of ways of selecting any 5 from total 24 = $24C5\\=42504$

No of ways of selecting only 5 girls = No of ways of selecting random 5 girls from total 12 girls

= $12C5\\=792$

a) the probability be that all five students

selected are girls= $\frac{792}{42504} \\=0.01863$

b) Since the probability for selecting all girls is very small and near to 0, it is unusual to select all girls if done at random. Hence the teacher had a preference for girls.

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3 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

sattari [20]

We know, that the <span>area of the surface generated by revolving the curve y about the x-axis is given by:

$\boxed{A=2\pi\cdot\int\limits_a^by\sqrt{1+\left(y'\right)^2}\, dx}$

In this case a = 0, b = 15, $y=\dfrac{x^3}{15}$ and:

$y'=\left(\dfrac{x^3}{15}\right)'=\dfrac{3x^2}{15}=\boxed{\dfrac{x^2}{5}}$

So there will be:

$A=2\pi\cdot\int\limits_0^{15}\dfrac{x^3}{15}\cdot\sqrt{1+\left(\dfrac{x^2}{5}\right)^2}\, dx=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\left(\star\right)\\\\-------------------------------\\\\
\int x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\,dx=\int\sqrt{1+\dfrac{x^4}{25}}\cdot x^3\,dx=\left|\begin{array}{c}t=1+\dfrac{x^4}{25}\\\\dt=\dfrac{4x^3}{25}\,dx\\\\\dfrac{25}{4}\,dt=x^3\,dx\end{array}\right|=\\\\\\$

$=\int\sqrt{t}\cdot\dfrac{25}{4}\,dt=\dfrac{25}{4}\int\sqrt{t}\,dt=\dfrac{25}{4}\int t^\frac{1}{2}\,dt=\dfrac{25}{4}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}= \dfrac{25}{4}\cdot\dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}=\\\\\\=\dfrac{25\cdot2}{4\cdot3}\,t^\frac{3}{2}=\boxed{\dfrac{25}{6}\,\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}}\\\\-------------------------------\\\\$

$\left(\star\right)=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\dfrac{2\pi}{15}\cdot\dfrac{25}{6}\cdot\left[\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}\right]_0^{15}=\\\\\\=
\dfrac{5\pi}{9}\left[\left(1+\dfrac{15^4}{25}\right)^\frac{3}{2}-\left(1+\dfrac{0^4}{25}\right)^\frac{3}{2}\right]=\dfrac{5\pi}{9}\left[2026^\frac{3}{2}-1^\frac{3}{2}\right]=\\\\\\=
\boxed{\dfrac{5\Big(2026^\frac{3}{2}-1\Big)}{9}\pi}$

Answer C.
</span>

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