Answer:
original size of Jay's bread dough is raised to 220%
Step-by-step explanation:
Let 1 be the original size of the bread
her bread dough is 11/5 of its original size.
The original size of the bread is raised to 2.2
To get percentage we multiply by 100
2.2*100 = 220%
original size of Jay's bread dough is raised to 220%
An equivalent name for the number 96 is word form ninety-six,or 90 + 6
Answer:
D. name an angle
Step-by-step explanation:
Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.
![~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20distance%20between%202%20points%7D%7D%7Bd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%7D~%5Chfill~%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B10%7D~%2C~%5Cstackrel%7By_1%7D%7B5%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B15%7D~%2C~%5Cstackrel%7By_2%7D%7B15%7D%29%20~%5Chfill%20a%3D%5Csqrt%7B%5B%2015-%2010%5D%5E2%20%2B%20%5B%2015-%205%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7Ba%3D%5Csqrt%7B125%7D%7D%20%5C%5C%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B15%7D~%2C~%5Cstackrel%7By_1%7D%7B15%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B30%7D~%2C~%5Cstackrel%7By_2%7D%7B9%7D%29%20~%5Chfill%20b%3D%5Csqrt%7B%5B%2030-%2015%5D%5E2%20%2B%20%5B%209-%2015%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7Bb%3D%5Csqrt%7B261%7D%7D)
![(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B30%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B10%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%20~%5Chfill%20c%3D%5Csqrt%7B%5B%2010-%2030%5D%5E2%20%2B%20%5B%205-%209%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7Bc%3D%5Csqrt%7B416%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctextit%7BHeron%27s%20area%20formula%7D%20%5C%5C%5C%5C%20A%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D%5Csqrt%7B125%7D%5C%5C%20b%3D%5Csqrt%7B261%7D%5C%5C%20c%3D%5Csqrt%7B416%7D%5C%5C%20s%5Capprox%2023.87%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%5Capprox%5Csqrt%7B23.87%2823.87-%5Csqrt%7B125%7D%29%2823.87-%5Csqrt%7B261%7D%29%2823.87-%5Csqrt%7B416%7D%29%7D%5Cimplies%20%5Cboxed%7BA%5Capprox%2090%7D)
SOH,CAH, TOA so it would be TANGENT since it would be Opposite over Adjacent. Tan(15) = 9/a put Tan(15) onto your calculator it would be .267949 round that to .27 = 9/a cross multiply
a = 33.588. C is the correct answer.
