Answer:
x = 21
Step-by-step explanation:
To see for what value of x is the equation true, we have to solve x.
![\displaystyle \frac{2}{3} x - 3 = 11\\\\Adding \ 3 \ to \ both \ sides\\\\\frac{2x}{3} = 11 + 3\\\\\frac{2x}{3} = 14\\\\Multiply \ 3 \ to \ both \ sides\\\\2x = 14 \times 3\\\\2x = 42 \\\\Divide\ both\ sides\ by\ 2\\\\x = 42/2\\\\x = 21\\\\For \ x = 21\ the \ given \ equation \ is \ true.\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B2%7D%7B3%7D%20x%20-%203%20%3D%2011%5C%5C%5C%5CAdding%20%5C%203%20%5C%20to%20%5C%20both%20%5C%20sides%5C%5C%5C%5C%5Cfrac%7B2x%7D%7B3%7D%20%3D%2011%20%2B%203%5C%5C%5C%5C%5Cfrac%7B2x%7D%7B3%7D%20%3D%2014%5C%5C%5C%5CMultiply%20%5C%203%20%5C%20to%20%5C%20both%20%5C%20sides%5C%5C%5C%5C2x%20%3D%2014%20%5Ctimes%203%5C%5C%5C%5C2x%20%3D%2042%20%5C%5C%5C%5CDivide%5C%20both%5C%20sides%5C%20by%5C%202%5C%5C%5C%5Cx%20%3D%2042%2F2%5C%5C%5C%5Cx%20%3D%2021%5C%5C%5C%5CFor%20%5C%20x%20%3D%2021%5C%20the%20%5C%20given%20%5C%20equation%20%5C%20is%20%5C%20true.%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3><h3>Peace!</h3>
Answer:
The interquartile range is 5.
Step-by-step explanation:
Ah, a throwback to interquartile range... let me help :)
4,5,6,8,9,10,11,12
First, you need to know how to use the IQR. The interquartile range is basically known as the process of subtracting the upper quartile and the lower quartile of a set of data. The lower quartile should be written as Q1, and the upper quartile would be labeled as Q3. This would make the midpoint (median) data set Q2, and the highest possible point would be labeled Q4. Next, you have to always understand what you are looking at. For example, let's split the set 5,6,7,8,9,10,11,12 into groups. 5 and 6 would be Q1, 7 and 8 would be Q2, 9 and 10 would be Q3, and last but not least, 11 and 12 would be labeled as Q4. Now take Q1 and subtract it from Q3 and that is how you get your IQR.
Answer:
-15 = n
Step-by-step explanation:
Answer:
Explanation:
The figure is not a regular hexagon. It is an irregular hexagon.
Please, find attached the picture with the original question and the figure.
You can split the figure into two triangles and one rectangle.
The rectangle has dimensions: 7units × 4units, thus its area is 28 units².
Both the upper triangle and lower triangle have base 7 units and height 2 units.
Hence the area of each triangle is:
Hence, the area of the hexagon is:
Radius is the distance between the center and any point on the circle.
r=distance =√((x2-x1)² +(y2-y1)²) =√((5-(-3))²+(-7-(-1))²)=√(64+36)=√100 = 10
r=√100
Equation of the circle , where (h, k) coordinates of the center of the radius.
(h,k)=(5,-7)
(x-h)² +(y-k)² = r²
(x-5)² +(y+7)² = (√100)²
(x-5)² +(y+7)² = 100
