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2 years ago

△ABC has vertices A(1,1) B(5,5) and C(4,6)

Mathematics

1 answer:

stira [4]2 years ago

5 0

Length in units:

Use phytogoras theorem or measure it by a ruler if your using a graph paper

AB = 5.65 units

AC = 5.83 units

CB = 1.41 units

Perimeter =5.65 + 5.83 + 1.41 = 12.89 units

Area = 1/2 * 1.41 * 5.65 = 3.98units

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Find the slope and the y-intercept of the following equation y = - x/3 - 4

BartSMP [9]

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The slope is -x/3 and the y-intercept is -4.

Step-by-step explanation:

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3 years ago

Val is going to plant y vegetable seeds in one garden 3y+5 vegetable seeds in another. Haw many seeds is val going to plant?

egoroff_w [7]

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4y+5 vegetable seeds.

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3 years ago

In your own words, explain to me what is absolute value? Make sure to talk about the definition of it, how you find it, and how

Tom [10]

Answer:

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2 years ago

How many different 10 person committees can be selected from a pool of 23 people?

Salsk061 [2.6K]

We're going to be using combination since this question is asking how many different combinations of 10 people can be selected from a set of 23.

We would only use permutation if the order of the people in the committee mattered, which it seems it doesn't.

Formula for combination:

$C(n,r)=\dfrac{n!}{(n-r)!r!}$

Where $n$ represents the number of objects/people in the set and $r$ represents the number of objects/people being chosen from the set

There are 23 people in the set and 10 people being chosen from the set

$C(23,10)=\dfrac{23!}{(23-10)!10!}$

$=\dfrac{23!}{13!\times10!}$

Usually I would prefer solving such fractions by hand instead of a calculator, but factorials can result in large numbers and there is too much multiplication. Using a calculator, we get

$=1,144,066$

Thus, there are 1,144,066 different 10 person committees that can be selected from a pool of 23 people. Let me know if you need any clarifications, thanks!

~ Padoru

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3 years ago

Read 2 more answers

What is the answer for this 2\5+m=5\6

Bingel [31]

Answer:

m=25/12

Step-by-step explanation:

multiply the reciprocal of 2/5 on both sides

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2 years ago