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Feliz [49]
3 years ago
10

Matthew is on the swim team every week he swims 20 miles if he practices 4 days each week and swims the same number of miles eac

h practice how many miles does he swim each day?
Mathematics
2 answers:
julsineya [31]3 years ago
5 0
This problem is one of division, and the setup is simple. All you would need to do is divide the number of miles Matthew swims by the number of days he practices.

20/4 = 5

Matthew swims 5 miles every day
iogann1982 [59]3 years ago
3 0
If Mathew swims an equal many miles for 4 days a week then to find out how many miles he swam each of those four days then you would divide. 20/4 which would give you your answer 5 miles
Hope this helps! :)
You might be interested in
Let g(x, y) = cos(x + 2y). (a) Evaluate g(2, -1). (b) Find the domain of g. (c) Find the range of g. 1. Find and sketch the doma
skelet666 [1.2K]

Not going to do (1), (2), (3) since they are implicit from (a) and (b) and require graphs (which aren't supported on brainly).

We have g(x,y)=\cos(x+2y), that means, if plug in x=2,y=-1 we get g(2,-1)=\cos(2+2(-1))=\cos(0)=\boxed{1}.

Doman of cosine is x\in[0,1].

Hope this helps.

4 0
3 years ago
Help help help thanks
Dmitry [639]

Answer: a=2/5b + 1/5c (I may have answered your question but idk if I did)

Step 1: Add -4a to both sides.

9a−2b+−4a=4a+c+−4a

5a−2b=c

Step 2: Add 2b to both sides.

5a−2b+2b=c+2b

5a=2b+c

Step 3: Divide both sides by 5.

5a/5=2b+c/5

a=2/5b+1/5c

7 0
3 years ago
Please help me with this it would be very much appreciated
lara [203]

Answer:

423.5

Step-by-step explanation:

12.1 times 10 times 7 = 847

847/2 = 423.5

6 0
3 years ago
Help me by please this is kinda hard for my child and I forgot it
lisabon 2012 [21]
It's 6!

When you multiply both the numerator and denominator by the same number, you will have an equivalent to the fraction you started with because the new fraction can be reduced to the original fraction.

2/3 = 4/6 = 6/9 = 8/12 = 10/15 = 12/15 etc.
4 0
3 years ago
15.
boyakko [2]

Explanation:

a)

{3 \choose 2} = \frac{3!}{2!(3-2)!} = \frac{6}{2*1} = 3

{3 \choose 1} = \frac{3!}{1!(3-1)!} = \frac{6}{1*2} = 3

b) First, we have that

{n \choose k} = \frac{n!}{k!(n-k)!}

On the other hand,

{n \choose n-k} = \frac{n!}{(n-k)!(n(n-k))!} = \frac{n!}{(n-k)!k!}

Therefore, both expressions are equal. This makes sense, because selecting k elements from a group of n is the same than specify which elements you will not select. In order to specify those elements you need to select the n-k elements that will not be selected. Hence, each time you are selecting k from n, you are also selecting n-k from n, and from that reason both combinatorial numbers are equal.

4 0
4 years ago
Read 2 more answers
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