I did (450*3/4=337.5) to get how much each person would eat, then I did (337.5*14=4725g) to see how big the turkey should be. I then converted grams to kilos (4725*0.001=4.725 kg), and I got how big the turkey would be. It turns out the appropriate size of the turkey would be medium, so then what I did was that I looked for the medium price of the turkey per kg, and found that it was £5.99 per kg. I did (4.725*5.99) and I got about £28.30.
It would cost £28.30 for 14 people.
Answer:
A 100 lb person will weight 40 lb on Mars
Step-by-step explanation:
Given:
Weight on Mars directly varies as its weight on earth.
Let weight on Mars be = 
Let weight on Earth be = 
Its known that:
∝ 
Thus
where
is constant of proportionality.
when
, then 
thus, we have
dividing both sides by 95.
∴ 
Thus when
,
would be calculated as


∴
(Answer)
∴ A 100 lb person will weight 40 lb on Mars.
Answer:
22
Step-by-step explanation:
We know that AC = 2 * AB so we can write:
12x - 4 = 2(5/2 x + 12)
12x - 4 = 5x + 24
7x = 28
x = 4
Since AB = BC the answer is 5/2 * 4 + 12 = 22.
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)
Using the
sine law.

The measure of side GH is
13.7 meters.