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3 years ago

How do I graph this?

Mathematics

1 answer:

OleMash [197]3 years ago

5 0

If you want to graph this you take the $500 plus $50 per hour to get the cost. the graph itself depends on what kind it is. (bar graph or a line graph)

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Help anyone can help me do the question,I will mark brainlest.

DIA [1.3K]

Answer:

Step-by-step explanation:

14.

by pythagoras,s theorem

$x=\sqrt{2^2+2^2} =\sqrt{8} =\sqrt{4 \times 2} =2\sqrt{2} \\y^2=x^2+2^2=(2\sqrt{2} )^2+2^2=8+4=12=4 \times 3\\y=2\sqrt{3}$

15.

first perpendicular=y(say)

and second perpendicular=z (say)

y²=9²+7²=81+49=130

z²=y²+5²=130+25=155

x²=z²+4²=155+16=171

x=√(171)

5 0

3 years ago

A house painter decides how many cans of paint to buy based on the total surface area of the house. The table below shows the su

makvit [3.9K]

Answer:

There is no image

Step-by-step explanation:

7 0

3 years ago

NO LINKS!!! Please help me with these problems

nadezda [96]

<h3>Answers:</h3>

7) Center= (-1,2) Radius= $\boldsymbol{\sqrt{8}}$ Equation: $(x+1)^2+(y-2)^2 = 8$

8) Center= (3,13) Radius= 13 Equation: $(x-3)^2+(y-13)^2 = 169$

=========================================================

Explanation:

Problem 7

Let's find the distance from (-1,2) to (-3,4)

$(x_1,y_1) = (-1,2) \text{ and } (x_2, y_2) = (-3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-1-(-3))^2 + (2-4)^2}\\\\d = \sqrt{(-1+3)^2 + (2-4)^2}\\\\d = \sqrt{(2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\$

This is the radius because it stretches from the center to a point on the circle, so $r = \sqrt{8}$

Squaring both sides will get us $r^2 = 8$

One useful template for a circle is the equation $(x-h)^2+(y-k)^2 = r^2\\\\$

(h,k) is the center

r is the radius

Let's plug in the given center (h,k) = (-1,2) and the r^2 value we found earlier.

$(x-h)^2+(y-k)^2 = r^2\\\\(x-(-1))^2+(y-2)^2 = 8\\\\(x+1)^2+(y-2)^2 = 8\\\\$

You can confirm this by using a tool like Desmos. See below.

------------------------------------------------------------------------

Problem 8

The endpoints of the diameter are (-2,1) and (8,25)

The center is the midpoint of these endpoints.

The midpoint of the x coordinates is (-2+8)/2 = 3

The midpoint of the y coordinates is (1+25)/2 = 13

The center is (h,k) = (3,13)

Now find the distance from the center to one of the points on the circle, let's say to (8,25)

$(x_1,y_1) = (3,13) \text{ and } (x_2, y_2) = (8,25)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-8)^2 + (13-25)^2}\\\\d = \sqrt{(-5)^2 + (-12)^2}\\\\d = \sqrt{25 + 144}\\\\d = \sqrt{169}\\\\d = 13\\\\$