All categories

3 years ago

Help with 8!!!!! Plz

Mathematics

1 answer:

weeeeeb [17]3 years ago

3 0

3(0) + 3 = 3
3(1) + 3 = 6
3(2) + 3 = 9
3(3) + 3 = 12
3(4)+3 = 15
Solve 3x+3 = 3
Subtract 3 from both sides so you get
3x = 0
X= 0

You might be interested in

Mary brought $20 to the skating rink. She paid $5.50 admission and $3.75 to rent skates. She also purchased two bottles of water

Salsk061 [2.6K]

Answer:

20-14.5=5.5

Step-by-step explanation:

7 0

3 years ago

Need help quick simplify the expression (-4) power of 3

sveticcg [70]

Answer:

-64

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

? - 2/3 = 1/2 what is the lost numer please

marta [7]

Step-by-step explanation:

Let x be the lost number then

x - 2/3 = 1/2

x = 1/2 + 2/3

x = (3 + 4)/6

x = 7/6

3 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)

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