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Burka [1]
3 years ago
10

What is the equation of the line that passes through the point (2, -1) and has a slope of ?​

Mathematics
1 answer:
DENIUS [597]3 years ago
7 0

Answer:

y = 3/2x - 4

Step-by-step explanation:

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Sales, property, and income are three types of _____. profits income interest taxes
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Answer should be taxes
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3 years ago
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Like terms 4x + 2x + 7x
nasty-shy [4]
Since the 3 of them are like terms, we have to ADD all 3 of them.

4x+2x+7x= 13x
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Pls help<br><br> Given: △ABC, CM⊥ AB, BC = 5, AB = 7<br> CA = 4 sqrt(2)<br> Find: CM
babymother [125]

Answer:

The general plan is to find BM and from that CM. You need 2 equations to do that.

Step One

Set up the two equations.

(7 - BM)^2 + CM^2 = (4*sqrt(2) ) ^ 2 = 32

BM^2 + CM^2 = 5^2 = 25

Step Two

Subtract the two equations.

(7 - BM)^2 + CM^2  = 32

BM^2 + CM^2         = 25

(7 - BM)^2 - BM^2 = 7               (3)

Step three

Expand the left side of the new equation labeled (3)

49 - 14BM + BM^2 - BM^2 = 7    

Step 4

Simplify And Solve

49 - 14BM = 7              Subtract 49 from both sides.

-49 - 14BM = 7 - 49

- 14BM = - 42              Divide by - 14

BM = -42 / - 14

BM = 3

Step  Five

Find CM

CM^2 + BM^2 = 5^2

CM^2 + 3^2 = 5^2        Subtract 3^2 from both sides.

CM^2 = 25 - 9            

CM^2 = 16                     Take the square root of both sides.        

sqrt(CM^2) = sqrt(16)

CM = 4    < Answer

Step-by-step explanation:

6 0
2 years ago
Solve for x. The triangles are similar.
Serjik [45]

Answer:

\fbox {D. 10}

Step-by-step explanation:

As the triangles are similar :

45/55 = (12x - 3)/143

9/11 = (12x - 3)/143

9 x 13 = 12x - 3

12x = 117 + 3

12x = 120

x = 10

5 0
2 years ago
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5. Find the general solution to y'''-y''+4y'-4y = 0
CaHeK987 [17]

For any equation,

a_ny^(n)+\dots+a_1y'+a_0y=0

assume solution of a form, e^{yt}

Which leads to,

(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0

Simplify to,

e^{yt}(y^3-y^2+4y-4)=0

Then find solutions,

\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}

For non repeated real root y, we have a form of,

y_1=c_1e^t

Following up,

For two non repeated complex roots y_2\neq y_3 where,

y_2=a+bi

and,

y_3=a-bi

the general solution has a form of,

y=e^{at}(c_2\cos(bt)+c_3\sin(bt))

Or in this case,

y=e^0(c_2\cos(2t)+c_3\sin(2t))

Now we just refine and get,

\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}

Hope this helps.

r3t40

5 0
3 years ago
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