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Doss [256]
3 years ago
8

PLEASE HELP ASAP

Mathematics
1 answer:
Igoryamba3 years ago
5 0

Answer:

∠1 - 40°

Step-by-step explanation:

∠1 - 40°

b/c it's a right triangle and we have two angles given, 50° and 90°. Add them and subtract by 180° and get 40°.

∠2 - 140°

b/c an exterior (outside) angle is equal to the two most isolated / farthest angles added. The two most is angles are 105° and 35°, add them and get 140°.

∠3 - 40°

b/c ∠'s 1 and 3 are vertical angles meaning they're equal so since ∠1 is 40°, so is ∠3.

∠4 -

b/c ∠' s 2 and 4 are vertical angles meaning they're equal so since ∠2 is 140°, so is ∠4.

∠5 - 35°

b/c we have two angles, 105° and 40°. Add them and subtract by 180° and get 35°.

~~

I hope that helps you out!!

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Answer:

Slope intercept form is y=mx+b

Let’s first find the slope, which is y2-y1/x2-x1

Let’s make the point (-2,-5) our x1 and y1 and let’s make the point (3,5) our x2 and y2.

5+5/3+2

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So far, our equation is looking like this:

y=2x + b

To find the y-intercept (b), let’s plug in one of the points.

Let’s use the point (3,5)

(5)=2(3) + b

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Step-by-step explanation:

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Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

4 0
3 years ago
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