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Allushta [10]
2 years ago
12

In the given diagram, which of the following pairs of angles are alternate interior angles?

Mathematics
2 answers:
denis-greek [22]2 years ago
7 0

Answer:

3 and 6, and 4 and 5

Step-by-step explanation:

They are the alternate interior angles. Hope this helps

castortr0y [4]2 years ago
3 0

Answer:

3 and 6

4 and 5

Step-by-step explanation:

Alternate interior angles are on the opposite sides of the transversal, t, and between the parallel lines, r and s

3 and 6 are alternate interior angles

4 and 5 are alternate interior angles

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According to the Venn diagram below, what is P(A∩B∩C)?
weqwewe [10]

6

The upside down U means intersection

A intersect B intersect C is where all 3 circles overlap.

Sum up all the numbers to get 50

P (a intersect b intersect c) = 6/50 = 3/25



3 0
3 years ago
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It takes 1 1/2 hours to fill 1/2 of a tank how long will it take to fill 2 tanks completely
Likurg_2 [28]

Answer:

15

Step-by-step explanation:

hope this helps

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2 years ago
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

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3 years ago
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nikklg [1K]

Answer:

C, 7 1/2

Step-by-step explanation:

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7 0
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Answer:

8-i=g

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