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Natasha2012 [34]
3 years ago
5

3 Write an equation of a line that passes through (-2,5) and has a slope of 4​

Mathematics
1 answer:
solmaris [256]3 years ago
6 0

Answer:

y = 4x + 13

Step-by-step explanation:

y = 4x + b

5 = 4(-2) + b

5 = -8 + b

13 = b

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Fill in the blanks to complete the description of the track. The track has ____ sides of the square and the distance around ____
rjkz [21]

Answer:

Step-by-step explanation:

The track has four sides of the square and the distance around 2 complete circle(s)

8 0
3 years ago
I NEED HELP PLEASE, THANKS :)
OLga [1]

A circle is made up of 360 degrees. Lets find the decimal form of each part of the circle.

White: 122.4/360 = 0.34

Red: 115.2/360 = 0.32

Silver: 68.4/360 = 0.19

Blue: 54/360 = 0.15

Now, solve for the car color that is not red or blue.

1 - (0.15 + 0.32)

1 - (0.47)

0.53

Therefore, the answer is C.

Best of Luck!

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3 years ago
What is the slope no games please i will report!
NNADVOKAT [17]
Here the slope is 2/6
8 0
3 years ago
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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n
GarryVolchara [31]

Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point (x,y)

Distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

Put (x,y)=(4,-2)

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

Put (x,y)=(2,-2)

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

7 0
3 years ago
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Murljashka [212]

Answer:

Step-by-step explanation:

sory I ain't know

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