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motikmotik
3 years ago
12

When you have an x value of zero in a point like this one (0,8) where is the point located on the graph? When you have a y value

of zero in a point like this ( -3,0) where is the point located on the graph?
plz explain
Mathematics
1 answer:
MArishka [77]3 years ago
3 0

Answer:

well I don't have a picture to explain it but (0,8) would be on the y axis and 8 on the x-axis ( it would go up 8 times and stay on the same line)

it would be -3 on the y axis and it wouldn't move up or down (move 3 points to the left)

Step-by-step explanation:

(x,y) when the x is 0 it doesn't move on the x-axis (left or right) and when it is 0 it doesn't move on the y-axis (up or down)

hope that makes sense!

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What is the average rate of change of g over the interval -8≤t≤2? Give an exact number.
soldi70 [24.7K]

Per the plot, looks like g(-8) = 1 and g(2) = -4. Then the average rate of change of g over [-8, 2] is

(g(2) - g(-8)) / (2 - (-8)) = (-3  - 1)/10 = -2/5

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3 years ago
Prove algebraically that the straight line with equation x =2y+5 is a tangent to the circle with equation x ² +y ²
mrs_skeptik [129]

The intersection point is only one. Then the equation of the line is tangent to the circle at point (1, -2).

<h3>What is a circle?</h3>

It is a locus of a point drawn an equidistant from the center. The distance from the center to the circumference is called the radius of the circle.

Prove algebraically that the straight line with equation x = 2y + 5 is a tangent to the circle with equation x² + y² = 5.

x = 2y + 5 ...1

x² + y² = 5 ...2

If the intersection of the point of the circle and line is one. Then the line is tangent to the circle.

Then from equations 1 and 2, we have

               (2y + 5)² + y² = 5

4y² + 25 + 20y + y² - 5 = 0

            5y² + 20y + 20 = 0

    5y² + 10y + 10y + 20 = 0

     5y (y + 2) + 10(y + 2) = 0

              (5y + 10)(y + 2) = 0

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Then the value of y is unique then the value of x will be unique.

The value of x will be

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The intersection point is only one. Then the equation of the line is tangent to the circle at point (1, -2).

More about the circle link is given below.

brainly.com/question/11833983

#SPJ1

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2 years ago
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3 years ago
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Substitute the solution from step 1 into the other equation.

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