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liubo4ka [24]
3 years ago
14

Which equation represents the line that passes through the points (3, 4) and (1, - 2)?

Mathematics
1 answer:
svlad2 [7]3 years ago
5 0
The answer is number 1, y= 3x -5
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Find the vertex points of y>-3x+1, 4y
hoa [83]
The vertex is in the form (x, y).

The x value of the vertex is found using x = -b/2a.

After finding the x values, plug it into the given inequality to find the y value of the vertex.

Then express the vertex as a point in the form (x, y).
4 0
3 years ago
-1/2=3/8y solve for y
koban [17]

Answer:

-4/3

Step-by-step explanation:

-\frac{1}{2} = \frac{3}{8}y\\-\frac{1}{2} \cdot \frac{8}{3} = \frac{3}{8}y \cdot \frac{8}{3} \\\\-\frac{8}{6} = y\\\boxed{y = -\frac{4}{3}.}

5 0
3 years ago
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If a= (3/6) and b=(-3/-5) find-3a+4b
iren2701 [21]

Answer:

-3a + 4b

= -3(3/6) + 4(-3/-5)

= -3/2 + 12/5

= 9/10

3 0
3 years ago
PLEASE HELP!!!
zmey [24]

Answer:

-2 1/3+ -11/6 = 4.1666666666666 repeating

Step-by-step explanation:

brainlest plz

6 0
3 years ago
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Find all the points having an x-coordinate of 4 whose distance from the point (-2,-1) is 10
Marta_Voda [28]
A(x_1;\ y_1);\ B(x_2;\ y_2)\\\\the\ distance\ between\ A\ and\ B:\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\A(-2;-1);\ B(4;\ y);\ d=10\\\\subtitute\\\\\sqrt{(4-(-2))^2+(y-(-1))^2}=10\\\\\sqrt{(4+2)^2+(y+1)^2}=10\\\\\sqrt{6^2+(y+1)^2}=10\\\\\sqrt{36+(y+1)^2}=10\ \ \ |square\ both\ sides\\\\36+(y+1)^2=10^2\\\\36+(y+1)^2=100\ \ \ |subtract\ 36\ from\ both\ sides\\\\(y+1)^2=64\iff y+1=-\sqrt{64}\ or\ y+1=\sqrt{64}\\\\y+1=-8\ or\ y+1=8\ \ \ \ |subtract\ 1\ from\ both\ sides\\\\y=-9\ or\ y=7

Answer:{\boxed{(4;-9)\ or\ (4;\ 7)}
4 0
3 years ago
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