Check the picture below.
since in a rhombus the diagonals bisect each other, thus EC = EA.
now, the rhombus is simply 4 congruent triangles, we know the base and height of one of them, thus
![\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=8\\ h=15 \end{cases}\implies A=\cfrac{1}{2}(8)(15)\implies A=60 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of all 4 triangles}}{4(60)\implies 240}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20triangle%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dbh~~%20%5Cbegin%7Bcases%7D%20b%3D8%5C%5C%20h%3D15%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%288%29%2815%29%5Cimplies%20A%3D60%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20all%204%20triangles%7D%7D%7B4%2860%29%5Cimplies%20240%7D)
Answer:
she runs .18 or 0.1785714286 miles per minute
Step-by-step explanation:
Answer:
B. G(x) = x³ - 5
Step-by-step explanation:
The graphs below are the same shape. They share the same formula but different y-intercepts.
F(x) = x³ has a y-intercept of 0.
F(x) = x³ + 0
G(x) has a y-intercept of -5.
G(x) = x³ - 5
So, we need to count up 1/4 each time till we get to 3.
1/4, 2/4, 3/4, 1, 1 1/4, 1 2/4, 1 3/4, 2, 2 1/4, 2 2/4, 2 3/4, 3.
Now, let's count how many times we counted to get to 3.
12.
So, Jill used 12 bottles of lotion.
Glad I could help, and good luck!
AnonymousGiantsFan
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