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3 years ago

TRIG PROOF! This was a bit tricky and has a huge weightage in my grade did i do it correctly

Mathematics

2 answers:

Eduardwww [97]3 years ago

8 0

you r correctoooo

gooooood luckkkkkkkkkkkkkkkkk

have a good daeee homieeeee

netineya [11]3 years ago

6 0

This is the correct answer. I agree with your friend.

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Find the area of the rectangle. Show your work!

Nesterboy [21]

3x^2 - 7x - 20

use FOILING

4 0

3 years ago

Systems By Graphing - Solve All Correctly For a Thanks, 5-Stars, And A Brianliest!

BartSMP [9]

Answer:

Step-by-step explanation:

Amari= Graph C and Solution is (-6,-2)

Bella= Graph A and Solution is (3,4)

Carl= Graph B and Solution is (0,-3)

0 solutions because the slopes are the same. The lines will never cross because they are at the exact same angle.

No, I do not agree with Joey because the lines have different slopes and will lead to the system cross which is the solution.

1st graph: y=2x-1 and y=7 solution #1= (4,7)

2nd graph: y=-2x-3 and y=1/2x solution #2= (-2,1)

3rd graph: y=x and y= -1/5+6 solution #3= (5,5)

I had this exact same assignment a few months ago, my teacher didn't use the 2nd slide but I had the 1st and 3rd slide so this should help!

4 0

3 years ago

What is 2/3 as a percent with one decimal place

Doss [256]

2/3 is .666666 so that would be the percent

3 0

3 years ago

Find Aut( 15). Use the Fundamental Theorem of Abelian Groups to express this group as an external direct product of cyclic group

Eduardwww [97]

Answer:

The solution is attached in the picture below

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

3 years ago