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Vaselesa [24]
4 years ago
7

Which function has the same why intercept as the graph below

Mathematics
1 answer:
damaskus [11]4 years ago
5 0
You mean 'y' intercept not 'why' intercept
y intercet means where the line crosses or intersepts the y axis

that also is when x=0
bsically, y intercept point is (0,yintercept)
in this graph , the y intercept is 3 units below the 0 so y intercept is -3 at point (0,-3)
y=mx+b
b=y intercept
we want to find an equation that is
y=mx-3
convert them into that

(1) y=(12-6x)/4
y=-3/2x+3
nope

(2)
27+3y=6x
subtract 27
3y=6x-27
divide 3
y=2x-9
nope

(3)
6y+x=18
subtract x
6y=-2x+18
'divide 6
y=-1/3x+3
nope

(4)
y+3=6x
subtract 3
y=6x-3
correct

answer is number 4
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Find the co-ordinates of the vertex of the function y=x²-12x+2 and show your work
nataly862011 [7]

Answer:

(6,-34)

Step-by-step explanation:

You can use the vertex formula to find the x of the vertex and then plugin the x value into the formula to find the y value.

x = -\frac{b}{2a}

x = -\frac{b}{2a}

x = -\frac{-12}{2(1)}

x = 6}

Now plugin 6 into  y=x²-12x+2

y=x²-12x+2

y=6²-12(6)+2

y=36-72+2

y=36-72+2

y = -34

Vertex = (6,-34)




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What is the length of one side of a die if the volumn of the die 27 cm^3
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a because the cubed root of 27 is 3

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5 0
4 years ago
4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia
Ad libitum [116K]

The <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its <em>standard</em> form from the coordinates of the foci and a <em>general</em> equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the <em>general</em> equation of the circle into its <em>standard</em> form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0​

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (<em>semi-major</em> distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

#SPJ1

5 0
2 years ago
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