Finding perimeter is simple. All you do is add up all of the side lengths, and bada bing bada boom, you got the perimeter.
Option A is correct.
Step-by-step explanation:
We need to identify which statement is true in matrix addition.
In matrix addition we just add the corresponding elements.
In the options given only Option A is correct i.e, (n+4)-7 = 18
All other options have error in them.
In Option B there should be -12 instead of +12
In Option C there should be -11 instead of 11
In Option D we are not adding corresponding entries.
So, Option A is correct.
Keywords: Matrix addition
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Answer:
It should be 1,628
.
Step-by-step explanation:
We see two shapes in the figure and those shapes are a rectangle and a circle. Let's find the area of the circle first. 25m won't help us find the area of the circle so let's pretend that 25m isn't there for now. 40m seems to be the diameter of the circle and to find the area of the circle we need to multiply the radius squared by Pi or 3.14. Half of 40 is 20 so we can multiply 20 squared by Pi to give us 1,256.63706. Do not round this number yet. As you can see this circle isn't a full circle. It's a semicircle. We can divide 1,256.63706 by 2 to find the area of a semicircle. You should get 628.31853 and round that to the nearest tenth and we get 628.32. Now let's keep that number in mind and find the area of the rectangle. Using a calculator we can easily multiply 40 and 25 to find the area of the rectangle, which is 1,000. Add 1,000 and 628.32 and our final results should be 1,628.32. I don't see this number in your options but option B is closely related to this answer. I hope this helps and if this answer is wrong then please give me some feedback on what I did wrong! Thank you!
I think the answer would be C
Check the picture below.
well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.
bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.
![\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}](https://tex.z-dn.net/?f=%5Cbf%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B1%7D-%5Cstackrel%7By1%7D%7B4%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B5%7D-%5Cunderset%7Bx_1%7D%7B1%7D%7D%7D%5Cimplies%20%5Ccfrac%7B-3%7D%7B4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bslope%20of%20AB%7D%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7B%5Cunderline%7Bnegative%20reciprocal%7D%20and%20slope%20of%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%7D%7B3%7D%7D)
so, it passes through the midpoint of AB,

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)
