Here we have some problems about circles, the solutions are:
- a) 490.6 yd^2
- b) x = 15
- c) x = 27
<h3>
How to find the area of a circle?</h3>
We know that for a circle of diameter D, the area is given by:
A = pi*(D/2)^2
where pi = 3.14
In the image we can see that the diameter of the circle is 25 yd, replacing that in the area equation we get:
A = 3.14*(25yd/2)^2 = 490.6 yd^2
<h3>How to find the value of x?</h3>
<u>For the second image</u>, each line creates two 180° angles. So, if we look at the tilted line, we can see that it is divided in two angles, one of 37° and the other that depends of x, so we can write:
180° = 37° + (9x + 8)°
180° - 37° - 8° = (9x)°
135°/9° = x = 15
So the value of x is 15.
For the last image:
Notice that we have a 90° angle denoted, then <u>the other 3 sections must add up to 270°.</u>
270° = 54° + (2x + 21)° + (5x + 6)°
Now we solve this for x:
270 - 54 - 21 - 6 = 2x + 5x
189 = 7x
189/7 = x = 27
x is equal to 27
If you want to learn more about circles, you can read:
brainly.com/question/25306774
Answer:
F(x) = 6x + 5.
Step-by-step explanation:
Given that :
Number of cards Given to Sébastian = 4
Number of baseball cards purchased every month = 6 cards
Number of months = x
Linear function f(x) which shows the number of cards after a number of months:
(Number of cards produced per month * Number of months) + number of card `
F(x) = 6x + 5
Answer:
1) 18a^11
2) 7b^5
3) 16c^12
4) 3/2d^2
Step-by-step explanation:
i hope this helped
Answer:
Shortest distance=|AC|/
.
Kindly find the attached for the figure
Step-by-step explanation:
This problem can be addressed using right-angled,
Let have right angle triangle with the shortest distance=hypotenuse;
hypotenuse=|AC|
using pythagoras theorem, we have
|AC|^2=|AB|^2+|BC|^2
Let |AB|=|BC|
|AC|^2=2|AB|^2
|AB|=|AC|/![\sqrt[2]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B2%5D%7B2%7D)
Shortest distance=|AC|/![\sqrt[2]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B2%5D%7B2%7D)
Answer:
j^13
Step-by-step explanation:
j j j j j j j j j j j j j j
There are 13 j's so we raise j to the 13 power
j^13