I neeeddd help pleaase

Mathematics

1 answer:

Norma-Jean [14]3 years ago

7 0

I got 65 for my answer. Hope that helps!

You might be interested in

Fundamental theorem of calculus <img src="https://tex.z-dn.net/?f=g%28s%29%3D%5Cint%5Climits%5Es_6%20%7B%28t-t%5E4%29%5E6%7D

mr_godi [17]

Answer:

$\displaystyle g'(s) = (s-s^4)^6$

Step-by-step explanation:

The Fundamental Theorem of Calculus states that:
$\displaystyle \frac{d}{dx}\left[ \int_a^x f(t)\, dt \right] = f(x)$

Where a is some constant.

We can let:
$\displaystyle g(t) = (t-t^4)^6$

By substitution:

$\displaystyle g(s) = \int_6^s g(t)\, dt$

Taking the derivative of both sides results in:
$\displaystyle g'(s) = \frac{d}{ds}\left[ \int_6^s g(t)\, dt\right]$

Hence, by the Fundamental Theorem:

$\displaystyle \begin{aligned} g'(s) & = g(s) \\ \\ & = (s-s^4)^6\end{aligned}$

3 0

1 year ago

If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

$\tan x$ is also positive for $x$ in quadrant I, so you take the positive root again. You're left with

$\tan\dfrac x2=\dfrac15$