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2 years ago

11

Cole builds a rocket with the dimensions shown. Suppose the rocket separates into the original parts before falling to Earth. Ho

w does the combined surface areas of the parts compare to the surface area of the assembled rocket?

2 answers:

Anettt [7]2 years ago

7 0

This was answered which is probably correct but.. Mind helping me on this? brainly.com/question/23506969 thanks!

agasfer [191]2 years ago

3 0

Answer:

In flight, a rocket is subjected to the forces of weight, thrust, and aerodynamics. On this slide, we have removed the outer "skin" so that we can see the parts. The payload of the German V2, shown in the figure, was several thousand

Step-by-step explanation:

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Please help me I’m so lost !!!!

Tasya [4]

Answer:

$4095$

Step-by-step explanation:

$\sum _{n=1}^63\left(4^{n-1}\right)\:$

$\sum _{n=1}^63\cdot \:4^{n-1}$

Compute general progression:

$\frac{3\cdot \:4^{\left(n+1\right)-1}}{3\cdot \:4^{n-1}}=4$

$r=4\\$

$a_1=3\cdot \:4^{1-1}$

$a_1=3$

nth term is computed by:

$r=4,\:a_n=3\cdot \:4^{n-1}$

plug in the values $n=6,\:\spacea_1=3,\:\spacer=4$ :

$=3\cdot \frac{1-4^6}{1-4}$

$=4095$

5 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

3 years ago

Expand the following x(x-6), x(4x-1), 2x(5x+4), 3x(5x-y)

zlopas [31]

Answer:

x^2-6x, 4x^2-1x,10x^2+8x,15x^2-3xy

Just use the distributive property like you have before and you will get these answers, Hope this helped!

4 0

3 years ago

Read 2 more answers

If NS bisects QNR find the measure of QNR

maksim [4K]

both angles would equal the same so you have:

6x-57 = 2x+15

add 57 to each side:

6x = 2x+72

subtract 2x from each side:

4x = 72

divide:

x = 72/4 = 18

now replace x with 18:

6(18) - 57 = 51

2(18) +15 = 51

51 +51 = 102

Angle QNR = 102 degrees

7 0

3 years ago

What is the slope of the line shown in the graph? (4 points)

Korvikt [17]

Answer:

a. or $\frac{3}{-2}$

Step-by-step explanation:

Slope is equal to y1-y2 over x1-x2.

So the equation would be:

$\frac{2-(-1)}{-3-(-1)}$

2-(-1)=2+1=3

-3-(-1)=-3+1=-2

So the answer is:

$\frac{3}{-2}$

5 0

2 years ago