NEED HELP ASAP

Mathematics

1 answer:

SIZIF [17.4K]2 years ago

4 0

Answer:

The answer is one solution because the graphs are together in one specific point.

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Find the interval(s) of upward concavity on this accumulation function.

maxonik [38]

$\displaystylef(x)=\int_{0}^{x^2}\sec^2(\sqrt{x})dx \\=\int_{0}^{\sqrt{x^2}}\sec^2(u)\cdot2udu \\=2\int_{0}^{\sqrt{x^2}}\sec^2(u)du \\=2\Big[u\tan(u)-\int\tan(u)du\Big]_{0}^{\sqrt{x^2}} \\=2\Big[u\tan(u)+\ln\Big(\mathrm{abs}(\cos(u))\Big)\Big]_0^x \\=2\Big(x\tan(x)+\dfrac{1}{2}\ln\Big(\cos^2(x)\Big)\Big) \\=2x\tan(x)+\ln(\cos^2(x))$

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6 0

3 years ago

A local high school has both male and female students.

Kitty [74]

Answer:

a. P(male) = 0.4

b. P(no sport and male) = 0.1

c. Unclear question (isn't it the same as b.?)

Step-by-step explanation:

The data below is what I've worked according to, which isn't very clear from the question so the answers are only correct if this is the correct table of data;

$\left[\begin{array}{ccc}&No \ Sports&Sports\\Female&10&32\\Male&7&21\end {array}\right]$