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Mashutka [201]
3 years ago
11

Solve the unknown variable: 5k/16 +2k/12 = (-25+k)/24

Mathematics
2 answers:
zvonat [6]3 years ago
8 0
5k/16+2k/12=(-25+k)/24
find common denomenator so we can cancel them
find GCM of 16,12, and 24

16=2*2*2*2
12=2*2*3
24=2*2*2*3
GCM=2*2*2*2*3=48

multiply everybody by 48

3(5k)+4(2k)=2(-25+k)
distribute
15k+8k=-50+2k
add like terms
23k=-50+2k
subtract 2k
21k=-50
divie both sides by 21
k=-50/21
k=-2 and 4/25
Goryan [66]3 years ago
6 0
Hello. (60k+32k)/192=(-25+k)/24 ; 92k/192=(-25+k)/24 ; 92k×24=192×(-25+k)/24 ; 2208k= -4800+192k ; 2208k-192k= -4800 ; 2016k= -4800 ; k= -4800/2016 ; k= -2,38. I hope to have helped you.
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A.Find a formula for
snow_lady [41]

Answer:

a) \frac{n}{n+1}

b) Proof in explanation.

Step-by-step explanation:

a)

\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)}.

So let's look at the last term for a minute:

\frac{1}{n(n+1)}

Let's use partial fractions to see if we can find a way to write this so it is more useful to us.

\frac{1}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}

Multiply both sides by n(n+1):

1=A(n+1)+Bn

Distribute:

1=An+A+Bn

Reorder:

1=An+Bn+A

Factor:

1=n(A+B)+A

This implies A=1 and A+B=0 which further implies that B=-1.

This means we are saying that:

\frac{1}{n(n+1)} can be written as \frac{1}{n}+\frac{-1}{n+1}

We can check by combing the fractions:

\frac{n+1}{n(n+1)}+\frac{-n}{n(n+1)}

\frac{n+1-n}{n(n+1)}

\frac{1}{n(n+1)}

So it does check out.

So let's rewrite our whole expression given to us using this:

(\frac{1}{1}+\frac{-1}{2})+(\frac{1}{2}+\frac{-1}{3})+(\frac{1}{3}+\frac{-1}{4})+\cdots +(\frac{1}{n}+\frac{-1}{n+1})

We should see that all the terms in between the first and last are being zeroed out.

That is, this sum is equal to:

\frac{1}{1}+\frac{-1}{n+1}

Multiply first fraction by (n+1)/(n+1) so we can combine the fractions:

\frac{n+1}{n+1}+\frac{-1}{n+1}

Combine fractions:

\frac{n}{n+1}

b)

Proof:

Let's see what happens when n=1.

Original expression gives us \frac{1}{1 \cdot 2}=\frac{1}{2}.

The expression we came up with gives us \frac{1}{1+1}=\frac{1}{2}.

So it is true for the base case.

Let's assume our expression and the expression given is true for some integer k greater than 1.

We want to now show it is true for integer k+1.

So under our assumption we have:

\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}=\frac{k}{k+1}

So let's add the (k+1)th term of the given series on both sides:

\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}

(Now we are just playing with right hand side to see if we can put it in the form our solution which be if we can \frac{k+1}{k+2}.)

I'm going to find a common denominator which will be (k+1)(k+2):

\frac{k}{k+1} \cdot \frac{k+2}{k+2}+\frac{1}{(k+1)(k+2)}

Combine the fractions:

\frac{k(k+2)+1}{(k+1)(k+2)}

Distribute:

\frac{k^2+2k+1}{(k+1)(k+2)}

Factor the numerator:

\frac{(k+1)^2}{(k+1)(k+2)}

Cancel a common factor of (k+1)

\frac{k+1}{k+2}

We have proven the given expression and our formula for the sum are equal for all natural numbers,n.

6 0
3 years ago
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