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elena-14-01-66 [18.8K]
3 years ago
10

What is the equation of the line that passes through the point (-5,7) and and has a slope of -2/5?

Mathematics
1 answer:
Vinvika [58]3 years ago
3 0

Answer:

y=-2/5 +7

Step-by-step explanation:

You might be interested in
Evaluate: 2/3 x 1<br> *Write your answer in fraction form.*
alina1380 [7]

Answer:

its just 2/3 lol

Step-by-step explanation:

2/3 x 1 is 2/3

8 0
3 years ago
How many ways can a gymnastics team of 4 be chosen from 9 gymnasts?
artcher [175]
You have 9 gymnasts and want to make a team of 4. Say one out of the nine is chosen; eight are left. Take one, then seven, and two more can be chosen. Keep doing this and your team is full. So, it's 9*8*7*6 which equals 3,024, since order doesn't matter.
3 0
3 years ago
Read 2 more answers
Harold found a puzzle with many pieces. He asked Carmen and Emma to help him solve the puzzle. He gave Carmen half his pieces. H
Leya [2.2K]

Answer:

72 pieces

Step-by-step explanation:

Basically split it up into

(1/6)+(1/6)+(1/6)+(1/6)+(1/6)+(1/6)

Ok so the bolded ones go to Carmen

(1/6)+(1/6)+(1/6)+(1/6)+(1/6)+(1/6)

so we are left with

(1/6)+(1/6)+(1/6)

the italics go to Emma

<em>(1/6)+(1/6)</em>+(1/6)

so now Harold only has 1/6 of the puzzle pieces which is 12 pieces.

Multiply 6 by 12 and you get 72

(Also Harold seems rude, giving so much work to his friends when its HIS puzzle) i swear you cant trust no one these days hmph

5 0
3 years ago
Can someone pls help me
Alina [70]

Answer:

Sure, what do you need?

4 0
3 years ago
A group of n students is assigned seats for each of two classes in the same classroom. how many ways can these seats be assigned
Ratling [72]
Consider ONLY the first class. Since there are n seats (assuming they are in a row), we would have n! ways.

Now, consider the second class. Let's start by arranging three people first and then generalising n people to better understand what is going on.

Where we have 3 people:
First class: A B C = 3!
Second class _ _ _
Now, consider where each of these people can't actually sit.
For A, it's the first seat, for B, it's the second seat, and for C, it's the last seat.
This means that we have a restriction on EACH of the candidates.

So, to tackle this, let's consider A only; B and C will follow the same way.
A can sit in 2 different spots, namely the second and last seat: thereby, having 3C2 ways in sitting. _ A _
Now, when we fix one person, C can only go in one place: first seat. This means that for one single arrangement of the first class, we've made: 3C2 arrangements for the second class, for ONE particular person. Extend that to another person, and we get: 2C1 ways

This extends to what we call: a derangement where the number of permutations made contains no fixed element. We can regard things like picking up three/two/one pen as derangements, because really we're arranging AND not arranging them simultaneously.

Thus, we use the inclusion-exclusion method:
Total no. of perms:
n! \cdot n!\left(1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - ... + (-1)^{n} \cdot \frac{1}{n!}\right)
4 0
3 years ago
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