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MissTica
3 years ago
11

Please Help! So Confused! Lines y and z are parallel. What is the measure of angle 2?

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
6 0
It's would be 65 degrees
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What is the answer to this question?#1 #2 #3 #4 #5 #6
Pavel [41]

Answer:

The solution is in the attached file

8 0
2 years ago
What is the area of the triangle 10 13 12
disa [49]

Answer:

A= 57 sq units

Step-by-step explanation:

To calculate the are area of a triangle we use the Herons formula:

A=√s(s-a)(s-b)(s-c) where a, b and c are the sides of the triangle.

s=(a+b+c)/2

s=(10+13+12)/2

=17.5

A=√[17.5(17.5-10)(17.5-13)(17.5-12)]

A= 57 sq units

5 0
3 years ago
What type of association is shown in the following scatter plot?
sergij07 [2.7K]
No association, because it’s they are all scattered
5 0
3 years ago
4(m+2) rewrite using distributive property
STALIN [3.7K]
4m + 8

Hope this helps!

4m + 8 = 0

Subtract 8 from both sides.

4m = -8

Divide 4 from both sides.

m = -2
3 0
3 years ago
Read 2 more answers
In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te^(−2t)−(8t+12) with
Zarrin [17]

First check the characteristic solution:

<em>y''</em> + 4<em>y'</em> + 4<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0

with a double root at <em>r</em> = -2, so the characteristic solution is

y_c = C_1e^{-2t} + C_2te^{-2t}

For the particular solution corresponding to 12te^{-2t}, we might first try the <em>ansatz</em>

y_p = (At+B)e^{-2t}

but e^{-2t} and te^{-2t} are already accounted for in the characteristic solution. So we instead use

y_p = (At^3+Bt^2)e^{-2t}

which has derivatives

{y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}

{y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}

Substituting these into the left side of the ODE gives

(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}

so that 6<em>A</em> = 12 and 2<em>B</em> = 0, or <em>A</em> = 2 and <em>B</em> = 0.

For the second solution corresponding to -8t-12, we use

y_p = Ct + D

with derivative

{y_p}' = C

{y_p}'' = 0

Substituting these gives

4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12

so that 4<em>C</em> = -8 and 4<em>C</em> + 4<em>D</em> = -12, or <em>C</em> = -2 and <em>D</em> = -1.

Then the general solution to the ODE is

y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1

Given the initial conditions <em>y</em> (0) = -2 and <em>y'</em> (0) = 1, we have

-2 = C_1 - 1 \implies C_1 = -1

1 = -2C_1 + C_2 - 2 \implies C_2 = 1

and so the particular solution satisfying these conditions is

y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1

or

\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}

7 0
2 years ago
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