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GalinKa [24]
3 years ago
10

Can you help me with this? thanks made it 20 points!

Mathematics
1 answer:
liraira [26]3 years ago
8 0
I agree with Roaltyjess<span />
You might be interested in
Point C ∈ AB and AB = 33 cm. Point C is 2 times farther from point B than point C is from point A. Find AC and CB.
forsale [732]

Answer:

AC = 11 cm , CB = 22 cm

Step-by-step explanation:

let AC = x then BC = 2x , then

AC + BC = 33, that is

x + 2x = 33

3x = 33 ( divide both sides by 3 )

x = 11

Thus

AC = x = 11 cm and CB = 2x = 2 × 11 = 22 cm

5 0
3 years ago
I run a book club with n people, not including myself. Every day, for 365 days, I invite three members in the club to review a b
Bezzdna [24]
<h3>Answer:   15</h3>

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

Explanation:

The order doesn't matter. A group like {A,B,C} is the same as {B,A,C}.

All that matters is the overall group rather than the positioning of the members.

We'll use the nCr combination formula since order doesn't matter.

The value of n is unknown, but we know that r = 3 members are to be selected.

Let's pick a value for n at random. Let's say n = 10.

Plug n = 10 and r = 3 into the nCr formula below.

n C r = \frac{n!}{r!(n-r)!}\\\\10 C 3 = \frac{10!}{3!*(10-3)!}\\\\10 C 3 = \frac{10!}{3!*7!}\\\\10 C 3 = \frac{10*9*8*7!}{3!*7!}\\\\ 10 C 3 = \frac{10*9*8}{3!}\\\\ 10 C 3 = \frac{10*9*8}{3*2*1}\\\\ 10 C 3 = \frac{720}{6}\\\\ 10 C 3 = 120\\\\

Unfortunately we don't reach 365 or larger.

Let's try n = 11

n C r = \frac{n!}{r!(n-r)!}\\\\11 C 3 = \frac{11!}{3!*(11-3)!}\\\\11 C 3 = \frac{11!}{3!*8!}\\\\11 C 3 = \frac{11*10*9*8!}{3!*8!}\\\\ 11 C 3 = \frac{11*10*9}{3!}\\\\ 11 C 3 = \frac{11*10*9}{3*2*1}\\\\ 11 C 3 = \frac{990}{6}\\\\ 11 C 3 = 165\\\\

We're still under our target. The good news is that the nCr value is increasing.

So the idea is to do trial and error with various values of n. Keep incrementing n until nCr = nC3 is equal to 365 or larger.

Here's a table of values where r = 3 the entire time

\begin{array}{|c|c|} \cline{1-2}\text{n} & \text{nCr}\\\cline{1-2}10 & 120\\\cline{1-2}11 & 165\\\cline{1-2}12 & 220\\\cline{1-2}13 & 286\\\cline{1-2}14 & 364\\\cline{1-2}15 & 455\\\cline{1-2}\end{array}

The nCr values are also found in Pascal's Triangle. Each of those values are the fourth entry of each row.

When n = 14, we have nCr = 364 which is very close. We're one short unfortunately.

So we have to go for <u>n = 15</u> instead. This makes the nCr value well over 365 of course, but it guarantees that you'll have plenty of trios to choose from such that no group of three is repeated. Unfortunately some trios will be left out.

7 0
2 years ago
What is .825 as a fraction and how do I find it
JulsSmile [24]

0.1 = \frac{1}{10}&#10;
0.01 = \frac{1}{100}&#10;&#10;
0.001 = \frac{1}{1000}

hence
0.825 = \frac{825}{1000}

now you just simplify the fraction
6 0
3 years ago
Given the functions k(x) = 2x2 − 7 and p(x) = x − 4, find (k ∘ p)(x).
Fynjy0 [20]

For this case we have the following functions:

k (x) = 2x ^ 2-7\\p (x) = x-4

We must find(k_ {0} p) (x). By definition of compound functions we have to:

(k_ {0} p) (x) = k (p (x))

So:

Taking into account that:

(a-b) ^ 2 = a ^ 2-2ab + b ^ 2

Different signs are subtracted and the major sign is placed.

k (p (x)) = 2 (x-4) ^ 2-7 = 2 (x ^ 2-2 (x) (4) + 4 ^ 2) -7 = 2 (x ^ 2-8x + 16) -7 = 2x ^ 2-16x + 32-7 = 2x ^ 2-16x + 25

Finally we have to:

(k_ {o} p) (x) = 2x ^ 2-16x + 25

Answer:

(k_ {o} p) (x) = 2x ^ 2-16x + 25

8 0
3 years ago
Who ever answers first gets brainiest!!
dlinn [17]

Answer:

10-B, 9-C, 11-D, 12-A

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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