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Lesechka [4]
2 years ago
7

In AWXY, mW Find mZX (2x – 12), m_X = (2x + 17), and m_Y = (9x-7) Find m

Mathematics
1 answer:
sladkih [1.3K]2 years ago
5 0

Answer:

45

Step-by-step explanation:

The sum of interior angles in a triangle is equal to 180

2x - 12 + 2x + 17 + 9x - 7 = 180 add like terms

13x - 2 = 180 add 2 to both sides

13x = 182 divide both side by 13

x = 14

m<X = 2x + 17 replace x with 14

2*14 + 17 = 45

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mash [69]
1. 1/3 cups..... ok vthen have a good day
4 0
3 years ago
50 points to whoever can answer!​
Pavlova-9 [17]

Answer:

D

Step-by-step explanation:

I think I'm not sure about that

4 0
3 years ago
Which of the following equations represents a line with a negative slope and a negative y-intercept?
Luden [163]
The answer is A because u can subtract the 3x to get 2y by itself then divide by 2 to completely get y by itself and it would be y=-3/2x - 9/2
5 0
3 years ago
On Friday a local hamburger shop sold a combined total of 564 hamburgers and cheeseburgers. The number of cheeseburgers sold was
Illusion [34]

Answer:

I believe it's 188

Step-by-step explanation:

If there was a total of 564 and the cheeseburgers was three times the amount, you need to divide 564 by 3. 564 divided by 3 equals to 188. I am pretty sure this could be the correct answer!

3 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
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