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aliina [53]
2 years ago
7

HELP HELP HELP HELP HELP HELP HELP

Mathematics
1 answer:
I am Lyosha [343]2 years ago
4 0
A) 3x (you substitute the g equation into the f one by putting it in the x value for f)

b) x (do the same thing, but switch the letters)
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Point C is located at (1, 2) and point D is located at (-4, -2). Find the X value for the point that is 1/4 the distance from po
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D (-4; -2)
C (1; 2)
M (x; y)
DM+MC=CD
CM:MD=1:3
M: x= x(C)-(x(C) - x(D))/4=1-(1-(-4))/4=1 - (6/4)=1 - 1,5= - 0,5
M: y=y(C)-(y(C) - y(D))/4=2-(2-(-2))/4=2 - (4/4)=2 - 1=1
M(x; y)=M( -0,5; 1)
8 0
3 years ago
Solve using quadratic equation by factoring.<br><br> x^2-9x=-14
Sonbull [250]

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Step-by-step explanation:

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5 0
3 years ago
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Rewrite the equation by completing the square.
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5 0
3 years ago
Volume8cm what is side lengths
pickupchik [31]

Answer:

length: 2

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Step-by-step explanation:

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6 0
3 years ago
Louisa ran at an average speed of five miles per hour along an entire circular park path. Calvin ran along the same path in the
docker41 [41]

Answer:

15 miles

Step-by-step explanation:

Let x be the miles in the circular park path, t_{L} the time Louisa takes to finish and t_{C} the time Calvin takes to finish both in hours.

Then x, the longitude is equal to the velocity times the time used to finish. So

x=5t_{L}

x=6t_{C}

And the difference between Louisa's time and Calvin' time is 30 minutes, half an hour. So:

t_{C}=t_{L}-0.5

Three equations, three unknowns, the system can be solved.

Equalizing the equation with x :

5t_{L}=6t_{C}

In this last equation replace t_{C}  with the other equation and solve:

5t_{L}=6(t_{L}-0.5)\\ 5t_{L}=6t_{L}-3\\ 3=6t_{L}-5t_{L}\\ 3=t_{L}\\ t_{L}=3

With Louisa's time find x:

x=5t_{L}\\ x=5(3)\\ x=15

7 0
3 years ago
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