Answer:
279 × 7 ≈ 2800
Step-by-step explanation:
if you have to round to the tens place, then both would be rounded up, the 7 to 10 and the 279 to 280. 10 × 280 = 2800
When any shape is inscribed in a circle, it means that the shape is within the circle but all of the corners are touching the circle. So this could just look like a square within a circle with the corners of the square touching the circle but not going outside of the borders of the circle. Repeat this process with the other shapes. The central angle used to locate the vertices is found by taking the number of sides on the shape and divide it by 360 (the angle of a circle). So for a square with 4 sides, you would take 360/4 and get 90 degrees. This means that each angle within the square is 90 degrees. That 90 degrees is the interior angle of the polygon (for a square specifically). Then what you do is look at the circle and draw a dot at the center of it. You can use a protractor for this part if you want but you would find the central angle by picking a point on the circle and drawing a line to the center dot, then you rotate however many degrees you found in the interior angle of the polygon and you would draw a new line from the center of the circle to that point. You will continue this process until you have gone back to your starting point on the circle. The amount of times it takes you to repeat the process should be the amount of sides the polygon has that you are trying. Interior angle and central angle should be the same for the individual shapes but it would be different for different shapes like a square and an octagon because there are a different amount of sides.
25 questions
Explanation: 20 divided by 25 is .8 which is 80%
Answer:
C
Step-by-step explanation:
well if you multiply each of tjem you would only sabe money on c the other choices would cause you to lose money so its c
If

is odd, then

while if

is even, then the sum would be

The latter case is easier to solve:

which means

.
In the odd case, instead of considering the above equation we can consider the partial sums. If

is odd, then the sum of the even integers between 1 and

would be

Now consider the partial sum up to the second-to-last term,

Subtracting this from the previous partial sum, we have

We're given that the sums must add to

, which means


But taking the differences now yields

and there is only one

for which

; namely,

. However, the sum of the even integers between 1 and 5 is

, whereas

. So there are no solutions to this over the odd integers.