Answer:
Step-by-step explanation:
A circle centered at (0,0) with radius r is 
Since your circle has diameter of 12, then its radius is 6. Then 
So a possible answer is:
If you want to move the location of the center, but keep it on the x-axis, then add or subtract a number to the x, such as this:
The center of this circle would be (-4, 0) which is still on the x-axis.
Answer:
<h2>10</h2>
Step-by-step explanation:
<em>The answer is 10 because 7 + 3 equals 10!</em>
<em></em>
<em>This can be proven by adding 7 + 3 or by subtracting in a certain way.</em>
<em>7 + 3 = ?</em>
<em>7 + 1 = 8</em>
<em>7 + 2 = 9</em>
<em>7 + 3 = 10</em>
<em></em>
<em>You can show more work by subtracting your answer or 10 in this case, which means that you subtract 7 by 10 or 3 by 10, either way is ok.</em>
<em></em>
<em>If you subtract 7 by 10, ( 10 - 7 ) then you get 3</em>
<em>If you subtract 3 by 10, ( 10 - 3 ) then you get 7</em>
<em>If the number you subtracted matches the number that you didn't subtract, then your answer is correct.</em>
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<em>Hope this helps! <3</em>
Answer:
2/3
Step-by-step explanation:
Benny had 60 cupcakes but only 4 tray to put them on, how many can go on each tray. Show your work
Let us formulate the independent equation that represents the problem. We let x be the cost for adult tickets and y be the cost for children tickets. All of the sales should equal to $20. Since each adult costs $4 and each child costs $2, the equation should be
4x + 2y = 20
There are two unknown but only one independent equation. We cannot solve an exact solution for this. One way to solve this is to state all the possibilities. Let's start by assigning values of x. The least value of x possible is 0. This is when no adults but only children bought the tickets.
When x=0,
4(0) + 2y = 20
y = 10
When x=1,
4(1) + 2y = 20
y = 8
When x=2,
4(2) + 2y = 20
y = 6
When x=3,
4(3) + 2y= 20
y = 4
When x = 4,
4(4) + 2y = 20
y = 2
When x = 5,
4(5) + 2y = 20
y = 0
When x = 6,
4(6) + 2y = 20
y = -2
A negative value for y is impossible. Therefore, the list of possible combination ends at x =5. To summarize, the combinations of adults and children tickets sold is tabulated below:
Number of adult tickets Number of children tickets
0 10
1 8
2 6
3 4
4 2
5 0
