Answer:
G. 7600 
Step-by-step explanation:
Step 1: Deconstruct the shape
2 rectangles with dimension of 50 cm x 20 cm
2 rectangles with dimension of 40 cm x 20 cm
2 rectangles with dimension of 40 cm x 50 cm
Step 2: Find surface area of the shapes
2(50 x 20) = 2000
2(40 x 20) = 1600
2(40 x 50) = 4000
Step 3: Add the surface areas together
2000 + 1600 + 4000 = 7600
Therefore the surface area is 7600 and the answer is G.
Answer:
50
Step-by-step explanation:
since they are parallel then 2x+6=106
so subtract 6 from each side so it's 2x = 100
then divide each since by 2
so x= 50
Answer:
a c and d
Step-by-step explanation:
Answer:
1.25 = X
Step-by-step explanation:
5-2x+2=12-4x-2x
Combine all like terms here.
7-2x=12-6x
Add 2x on both sides.
7=12-4x
Subtract 12 on both sides.
-5=-4x
Divide by -4.
1.25=X
Check the picture below.
since the vertical distance, namely the y-coordinate, is twice as much as the horizontal, then if the horizontal is "x", the vertical one must be 2x.
let's find the hypotenuse first.
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{2x}\\ \end{cases} \\\\\\ c=\sqrt{x^2+(2x)^2}\implies c=\sqrt{x^2+4x^2}\implies c=\sqrt{5x^2}\implies c=x\sqrt{5} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3D%5Cstackrel%7Badjacent%7D%7Bx%7D%5C%5C%20b%3D%5Cstackrel%7Bopposite%7D%7B2x%7D%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7Bx%5E2%2B%282x%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7Bx%5E2%2B4x%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B5x%5E2%7D%5Cimplies%20c%3Dx%5Csqrt%7B5%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{2~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}{\stackrel{hypotenuse}{~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ \sqrt{5}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \cfrac{2\sqrt{5}}{5}}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B2~~%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B~~%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%5Csqrt%7B5%7D%7D%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Band%20rationalizing%20the%20denominator%7D~%5Chfill%20%7D%7B%5Ccfrac%7B2%7D%7B%5Csqrt%7B5%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%7B5%7D%7D%7B%5Csqrt%7B5%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%7B5%7D%7D%7B%28%5Csqrt%7B5%7D%29%5E2%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%7B5%7D%7D%7B5%7D%7D)
