Yes, it is because there is 100 mL in 1 L
100 mL=1 L
Multiply the whole equation by 4.
400 mL=4 L
Hope this helps :)
we have that

I. Rewrite the equation by substituting the expression u in for sin x.

II. Factor the quadratic expression. Rewrite the equation with factors instead of the original polynomial.
is equal to
using a graph calculator-----> see the attached figure

III. Use the zero product property to solve the quadratic equation.

(u-3)=0--------------> u=3
(2u+1)=0-------- 2u=-1------> u=-1/2-----> u=-0.5
IV. Rewrite your solutions to Part III by replacing u with sin x.
sin x=3--------> is not the solution (sin x can not be greater than 1)
sin x=-0.50------>is the solution
V. Solve the remaining equations for x, giving all solutions to the equation.
sin x=-0.50
if the sine is negative
then
x belong to the III or IV quadrant
we know that
sin 30°=0.50
so
the solution for the III quadrant is
x=180°+30°-------> x=210°
the solution for the IV quadrant is
x=360°-30°------> x=330°
Answer:
x = -8
Step-by-step explanation:
I will show each step algebraically. If you need further explanation, feel free to ask questions :)
Rewrite
Subtract 7
Multiply by reciprocal (2/1)
I hope this helps!
Check the picture below.
![\stackrel{\textit{\Large Areas}}{\stackrel{triangle}{\cfrac{1}{2}(6)(6)}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}\pi (3)^2}}\implies \boxed{18+4.5\pi} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{pythagorean~theorem}{CA^2 = AB^2 + BC^2\implies} CA=\sqrt{AB^2 + BC^2} \\\\\\ CA=\sqrt{6^2+6^2}\implies CA=\sqrt{6^2(1+1)}\implies CA=6\sqrt{2} \\\\\\ \stackrel{\textit{\Large Perimeters}}{\stackrel{triangle}{(6+6\sqrt{2})}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}2\pi (3)}}\implies \boxed{6+6\sqrt{2}+3\pi}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%286%29%286%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20%283%29%5E2%7D%7D%5Cimplies%20%5Cboxed%7B18%2B4.5%5Cpi%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bpythagorean~theorem%7D%7BCA%5E2%20%3D%20AB%5E2%20%2B%20BC%5E2%5Cimplies%7D%20CA%3D%5Csqrt%7BAB%5E2%20%2B%20BC%5E2%7D%20%5C%5C%5C%5C%5C%5C%20CA%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B6%5E2%281%2B1%29%7D%5Cimplies%20CA%3D6%5Csqrt%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Perimeters%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%286%2B6%5Csqrt%7B2%7D%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D2%5Cpi%20%283%29%7D%7D%5Cimplies%20%5Cboxed%7B6%2B6%5Csqrt%7B2%7D%2B3%5Cpi%7D)
notice that for the perimeter we didn't include the segment BC, because the perimeter of a figure is simply the outer borders.