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a_sh-v [17]
3 years ago
5

Find the area of the triangle defined by the coordinates (-3,-4), (-7.0), and (-3, 4).

Mathematics
1 answer:
likoan [24]3 years ago
5 0

i think D is the answer 16

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2 years ago
What is the approximate solution to the equation 2^t =38
timurjin [86]

2^t =38

take the log of each side

log ( 2^t) = log (38)

the exponent gets multiplies

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t = log(38)/log (2)

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3 0
3 years ago
Please answer this question now
lidiya [134]

Answer:

C = (2,2)

Step-by-step explanation:

B = (10 ; 2)

M = (6 ; 2)

C = (x ; y )

|___________|___________|

B (10;2)            M (6;2)             C ( x; y)

So:

dBM = dMC

√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 6)^2]

(2-2)^2 - (6-10)^2 = (y-2)^2 + (x - 6)^2

0 + (-4)^2 = (y-2)^2 + (x - 6)^2

16 = (y-2)^2 + (x - 6)^2

16 - (x - 6)^2 = (y-2)^2

Also:

2*dBM = dBC

2*√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 10)^2]

4*[(0)^2 + (-4)^2] = (y-2)^2 + (x - 10)^2

4*(16) = (y-2)^2 + (x - 10)^2

64 = (y-2)^2 + (x - 10)^2

64 = 16 - (x - 6)^2 + (x - 10)^2

48 = (x - 10)^2 - (x - 6)^2

48 = x^2 - 20*x + 100 - x^2 + 12*x - 36

48 = - 20*x + 100 + 12*x - 36

8*x = 16

x = 2

Thus:

16 - (x - 6)^2 = (y-2)^2

16 - (2 - 6)^2 = (y-2)^2

16 - (-4)^2 = (y-2)^2

16 - 16 = (y-2)^2

0 =  (y-2)^2

0 = y - 2

2 = y

⇒ C = (2,2)

4 0
3 years ago
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