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arsen [322]
3 years ago
5

I need urgent help plz

Mathematics
2 answers:
WITCHER [35]3 years ago
7 0
C. And E.
The other three choices have outliers
Katarina [22]3 years ago
3 0

Answer:

c e

Step-by-step explanation:

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3 years ago
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Find the length of the indicated side in each triangle, to nearest unit.
gogolik [260]

Answer:

A= 56

E= 59

Step-by-step explanation:

All triangle are 180 degrees so

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71 + 53 = 124

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70 + 51 = 121

180 - 121 = 59

Hope this helps

4 0
2 years ago
What is the period of the graph y=1/2sin (pix)+3?<br><br> A- 3<br> B- 1/2<br> C-pi <br> D-2
dimaraw [331]

The period is 2.

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3 0
3 years ago
What is the domain and the range of these functions?? quick please!!!!!!!!!!
Sever21 [200]
The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes. A rational function is a function of the form f(x)=p(x)q(x) , where p(x) and q(x) are polynomials and q(x)≠0 .
6 0
3 years ago
Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)
vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\

Apply substitution to get

y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right)  + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x)  + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x)  + \left(a - 6b\right)\cos(x) = \sin(x)\\\\

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a  -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0
2 years ago
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