Check the picture below.
so hmmm let's find the distance AC, which is the only one missing for the perimeter.
![\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2 + b^2} \qquad \begin{cases} c=\stackrel{hypotenuse}{AC}\\ a=\stackrel{adjacent}{6}\\ b=\stackrel{opposite}{8}\\ \end{cases} \implies AC=\sqrt{6^2 + 8^2} \\\\\\ AC=\sqrt{100}\implies AC=10~\hfill \underset{Perimeter}{\stackrel{6~~ + ~~8~~ + ~~10}{\text{\LARGE 24}cm}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%20%2B%20b%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3D%5Cstackrel%7Bhypotenuse%7D%7BAC%7D%5C%5C%20a%3D%5Cstackrel%7Badjacent%7D%7B6%7D%5C%5C%20b%3D%5Cstackrel%7Bopposite%7D%7B8%7D%5C%5C%20%5Cend%7Bcases%7D%20%5Cimplies%20AC%3D%5Csqrt%7B6%5E2%20%2B%208%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AC%3D%5Csqrt%7B100%7D%5Cimplies%20AC%3D10~%5Chfill%20%5Cunderset%7BPerimeter%7D%7B%5Cstackrel%7B6~~%20%2B%20~~8~~%20%2B%20~~10%7D%7B%5Ctext%7B%5CLARGE%2024%7Dcm%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ b=6\\ h=8 \end{cases}\implies A=\cfrac{1}{2}(6)(8)\implies A=\text{\LARGE 24}~cm^2](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20triangle%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dbh%20%5Cbegin%7Bcases%7D%20b%3Dbase%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20b%3D6%5C%5C%20h%3D8%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%286%29%288%29%5Cimplies%20A%3D%5Ctext%7B%5CLARGE%2024%7D~cm%5E2)
Plug the y value in
2(4)^3
Evaluate exponents first (order of operations)
2(64)
Multiply
128
Final answer: 128
What's are you trying to find?
Answer:
Total number of tables of first type = 23.
Total number of tables of second type = 7
Step-by-step explanation:
It is given that there are 30 tables in total and there are two types of tables.
Let's call the two seat tables, the first type as x and the second type as y.
∴ x + y = 30 ......(1)
Also a total number of 81 people are seated. Therefore, 2x number of people would be seated on the the first type and 5y on the second type. Hence the equation becomes:
2x + 5y = 81 .....(2)
To solve (1) & (2) Multiply (1) by 2 and subtract, we get:
y = 7
Substituting y = 7 in (1), we get x = 23.
∴ The number of tables of first kind = 23
The number of tables of second kind = 7
Answer:
2000 Pounds
Step-by-step explanation:
There are 2000 pounds in a US ton.
There are 2240 pounds in an Imperial ton.
There are 2204.62 pounds in a metric ton.
