Help me please......

How to find the derivative of an integral with bounds.

Free_Kalibri [48]

Answer:

The statement of the fundamental theorem of calculus shows the upper limit of the integral as exactly the variable of differentiation. Using the chain rule in combination with the fundamental theorem of calculus we may find derivatives of integrals for which one or the other limit of integration is a function of the variable of differentiation.

Step-by-step explanation:

3 0

2 years ago

Rewrite in simplest radical form square root of x times the fourth root of x. Show each step of your process.

Ipatiy [6.2K]

Answer:

$\sqrt[4] {x^3}$

Step-by-step explanation:

At this point, we can transform the square root into a fourth root by squaring the argument, and bring into the other root:

$\sqrt x \cdot \sqrt[4] x =\sqrt [4] {x^2} \cdot \sqrt[4] x = \sqrt[4]{x^2\cdot x} = \sqrt[4] {x^3}$

Alternatively, if you're allowed to use rational exponents, we can convert everything:

$\sqrt x \cdot \sqrt[4] x = x^{\frac12} \cdot x^\frac14 = x^{\frac12 +\frac14}= x^{\frac24 +\frac14}= x^\frac34 = \sqrt[4] {x^3}$

5 0

1 year ago

Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )

Zina [86]

Answer:

$\purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ \sqrt{6} }{6} = \frac{1}{ \sqrt{6} } \\ \\ \therefore \: \sin( \alpha ) = \sqrt{1 - { \cos}^{2} ( \alpha ) } \\ \\ = \sqrt{1 - \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} } \\ \\ = \sqrt{1 - {\frac{1}{ {6} }}} \\ \\ = \sqrt{ {\frac{6 - 1}{ {6} }}} \\ \\ \red{\sin( \alpha ) = \sqrt{ { \frac{5}{ {6} }}} } \\ \\ \tan( \alpha ) = \frac{\sin( \alpha ) }{\cos( \alpha ) } = \sqrt{5} \\ \\ \sin( \beta ) = \frac{ \sqrt{2} }{4} \\ \\ \implies \: \cos( \beta ) = \sqrt{ \frac{7}{8} } \\ \\ \tan( \beta ) = \frac{\sin( \beta ) }{\cos( \beta ) } = \frac{1}{ \sqrt{7} } \\ \\ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha . \tan \beta} \\ \\ = \frac{ \sqrt{5} - \frac{1}{ \sqrt{7} } }{1 + \sqrt{5} . \frac{1}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{35} - 1 }{ \sqrt{7} + \sqrt{5} } \\ \\ \purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

8 0

3 years ago

Keith as 11/12 hours to play, he has already played 1/4. which is the best estimate of the fractional part of an hour he has lef

____ [38]

Answer:

The best estimate is greater than 1/2 but less than 3/4

Step-by-step explanation:

If you multiply 1/4 by 3/3 (which is also 1), you can change the denominator without changing the value. So 1/4 is equal to 3/12.

Since keith has 11/12 hours to play, and he has already played 3/12 hours, subtract 3/12 from 11/12 to get 8/12 hours. This is how much time he has left to play.

If you simplify 8/12 hours, you get 2/3 hours.

So the best estimate would be: greater than 1/2 but less than 3/4.

4 0

3 years ago

Using the diagram, what is the intersection of plane ABCD and ? A. Line B. Point C C. Plane EFGH D. Point D

GuDViN [60]

I have attached screenshots for the complete question as well as the diagram associated with it.

Answer:
point C

Explanation:
A plane and a line can either:
1- have no intersection
2- intersect in a point
3- intersect in a line in case the whole line is found in this plane

For the given:
We can see that line CG does not belong to the plane ABCD, however, they both intersect.
This means that the intersection between them is a point.
We can note that the only point common between the line and the plane is point C.
Therefore, plane ABCD intersects line GC at point C

Hope this helps :)

3 0

3 years ago

Help me please......​

Help me please......