We would divide the shape into a rectangle and a trapezium
The rectangle is of length, 24.5 ft and width 10.75 ft
Area = 24.5 * 10.75 = 263.375 square feet.
Trapezium : Parallel sides are 18.5 ft and 24.5 ft
Height: 12 - 10.75 = 1.25ft
Area of Trapezium = (1/2)*(Sum of Parallel sides)*height
= 1/2 * (18.5+24.5)* 1.25
= 1/2 * 43* 1.25 = 26.875
Total Area = Area of Rectangle + Area of Trapezium
= 263.375 + 26.875 = 290.25
Area = 290.25 ft² (D)
Hello!
We are given the weight of Mia's dog and told to find the weight of Lettie's dog based on the known relationship. Using these given values, we can create the following equation (let "w" represent the total weight of Lettie's dog):
w = 32.6(3.8)
Simplify the right side of the equation:
w = 123.88
We have now proven that Lettie's dog weighs 123.88 lbs.
I hope this helps!
Answer: true
Step-by-step explanation:
Divide centimeters by 100. Answer is 1.52 meters.
let's first off take a peek at those values.
let's say the point with those coordinates is point C, so C is 3/10 of the way from A to B.
meaning, we take the segment AB and cut it in 10 equal pieces, AC takes 3 pieces, and CB takes 7 pieces, namely AC and CB are at a 3:7 ratio.
![\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(-4,-8)\qquad B(11,7)\qquad \qquad \stackrel{\textit{ratio from A to B}}{3:7} \\\\\\ \cfrac{A\underline{C}}{\underline{C} B} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(-4,-8)=3(11,7)\\\\[-0.35em] ~\dotfill\\\\ C=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Binternal%20division%20of%20a%20line%20segment%7D%0A%5C%5C%5C%5C%5C%5C%0AA%28-4%2C-8%29%5Cqquad%20B%2811%2C7%29%5Cqquad%0A%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bratio%20from%20A%20to%20B%7D%7D%7B3%3A7%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BA%5Cunderline%7BC%7D%7D%7B%5Cunderline%7BC%7D%20B%7D%20%3D%20%5Ccfrac%7B3%7D%7B7%7D%5Cimplies%20%5Ccfrac%7BA%7D%7BB%7D%20%3D%20%5Ccfrac%7B3%7D%7B7%7D%5Cimplies%207A%3D3B%5Cimplies%207%28-4%2C-8%29%3D3%2811%2C7%29%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0AC%3D%5Cleft%28%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22x%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cquad%20%2C%5Cquad%20%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22y%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cright%29%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill)
![\bf C=\left(\cfrac{(7\cdot -4)+(3\cdot 11)}{3+7}\quad ,\quad \cfrac{(7\cdot -8)+(3\cdot 7)}{3+7}\right) \\\\\\ C=\left( \cfrac{-28+33}{10}~~,~~\cfrac{-56+21}{10} \right)\implies C=\left( \cfrac{5}{10}~~,~~\cfrac{-35}{10} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill C=\left( \frac{1}{2}~,~-\frac{7}{2} \right)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20C%3D%5Cleft%28%5Ccfrac%7B%287%5Ccdot%20-4%29%2B%283%5Ccdot%2011%29%7D%7B3%2B7%7D%5Cquad%20%2C%5Cquad%20%5Ccfrac%7B%287%5Ccdot%20-8%29%2B%283%5Ccdot%207%29%7D%7B3%2B7%7D%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0AC%3D%5Cleft%28%20%5Ccfrac%7B-28%2B33%7D%7B10%7D~~%2C~~%5Ccfrac%7B-56%2B21%7D%7B10%7D%20%5Cright%29%5Cimplies%20C%3D%5Cleft%28%20%5Ccfrac%7B5%7D%7B10%7D~~%2C~~%5Ccfrac%7B-35%7D%7B10%7D%20%5Cright%29%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20C%3D%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D~%2C~-%5Cfrac%7B7%7D%7B2%7D%20%5Cright%29~%5Chfill)
