What is the area of a regular pentagon if its apothem has a length of 4 feet and each side has a length of 5.8 feet?

A car travels along a straight road for 30 seconds starting at time t = 0. Its acceleration in ft/sec2 is given by the linear gr

I am Lyosha [343]

From an acceleration-time graph, you can work out the area under the graph to find the velocity. There are two triangles on the graph as shown below. The total area of the triangles are

$\frac{(10)(10)}{2}+ \frac{(20)(20)}{2}=250 ft/sec^{2}$

Hence, the distance is 250×30=7500 ft

8 0

3 years ago

Helpppp me guysssss thanksssss

castortr0y [4]

Answer:

help u with wat??

Step-by-step explanation:

4 0

3 years ago

PLEASE HELP!!!

Allushta [10]

Answer:

- 1/2 7/12 2/3 . 82

Am I right

8 0

3 years ago

A sector of a circle has a central angle of 60°. find the area of the sector if the radius of the circle is 6 mi.

professor190 [17]

The entire circle has a central angle of 360°.
Because the radius is 6 mi, the area of the circle is
π(6)² = 36π mi².

The sector has a central angle of 60°, therefore it occupies 60/360 = 1/6 of the area of the circle.
The area of the sector is
(1/6)*36π = 6π mi² = 18.85 mi²

Answer: 6π mi² or 18.85 mi².

8 0

3 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi

stealth61 [152]

Applying our power rule gets us our first derivative,

$\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}$

simplifying a little bit,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

looking for critical points,

$\rm 0=2x^{-2/3}+4x^{1/3}$

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

$0=2x^{-2/3}\left(1+2x\right)$

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

$\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)$

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

and take our second derivative.

$\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Looking for inflection points,

$\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Again, pulling out the smaller power of x, and fractional part,

$\rm 0=-\frac43x^{-5/3}\left(1-x\right)$

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.

8 0

4 years ago