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monitta
2 years ago
11

I’ll give brainliest help a girl out

Mathematics
2 answers:
never [62]2 years ago
4 0

Answer:

ST, WX

Step-by-step explanation:

Lady bird [3.3K]2 years ago
3 0

Answer:

ST and WX are congruent

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Factoring the expression 15x+6 using GCF
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The answer is 3(5x+2) i hope this helps you.
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Brittany is at a restaurant. Her bill is $48 plus 15%. She gave an $100. How much did she get back?
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She got 44.80 dollars back
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Read 2 more answers
Write 3,272 in expanded form
rosijanka [135]
3272 I 2
1636 I 2
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  409 I 409    so bc. 409 is a prime number has factors of 409 and 1 
      1 I

so the expanded form of 3272 will be 2^3 *409

hope helped 
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3 years ago
Which matrix translates the vector 〈4,-3〉 to 〈6,-8〉?
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Step-by-step explanation:

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3 years ago
(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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