<img src="https://tex.z-dn.net/?f=%5Ccolor%7Bred%7D%7B%5Ctt%20%7BIn%20%5C%3A%20%20%CE%94%20%5C%3A%20%20ABC%2C%20%E2%88%A0B%3D90%

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2 years ago

2C%20AB%20%3D%208%20%5C%3A%20%20cm%20%20%5C%3A%20and%20%5C%3A%20%20BC%20%3D%206%20%20%5C%3A%20cm.%20Find%20%5C%3A%20%20the%20%5C%3A%20%20length%20%5C%3A%20%20of%20%5C%3A%20%20AC.%20%7D%7D" id="TexFormula1" title="\color{red}{\tt {In \: Δ \: ABC, ∠B=90, AB = 8 \: cm \: and \: BC = 6 \: cm. Find \: the \: length \: of \: AC. }}" alt="\color{red}{\tt {In \: Δ \: ABC, ∠B=90, AB = 8 \: cm \: and \: BC = 6 \: cm. Find \: the \: length \: of \: AC. }}" align="absmiddle" class="latex-formula">

Mathematics

2 answers:

Sliva [168]2 years ago

6 0

Answer:

Solution given;

<B=90

AB=b=8cm

BC=p=6cm

AC=h=?

by using Pythagoras law

h²=p²+b²

h=√{6²+8²)

h=10cm

length of AC=10cm.

makkiz [27]2 years ago

5 0

Use Pythagoras theorem

Hypotenuse²=Base²+ Height ²

H²=(8)²+(6)²

H²=64+36

H²= 910

H²=10√91 cm Solution

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How to calculate the diameter of cylinder if you are given volume of 1000Litre and 224cm height

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1 year ago

Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size a

Nikitich [7]

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Let the side lengths of the rectangular pan be m and n. It follows that (m-2) (n-2)= $mn/2$

So, since haf of the brownie pieces are in the interior. This gives 2 (m-2) (n-2) =mn

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3 years ago

7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many

BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

$_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\$

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

$_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\$

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

$_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\$

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>

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2 years ago