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Arlecino [84]
3 years ago
15

How do i find the least common denominator of 3/8 and 5/6?No links please.​

Mathematics
1 answer:
torisob [31]3 years ago
8 0

Step-by-step explanation:

first find a number that makes divided both if you don't get multiply the denominators

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4x-5 - 2x-1
3241004551 [841]

Answer:

\frac{5x - 8}{3}

Step-by-step explanation:

Find the LCM of the denominators and then solve:

\frac{4x - 5}{2} - \frac{2x - 1}{6} \\\\

\frac{4x - 5}{2} - \frac{2x - 1}{6} \\\\\frac{3(4x - 5) - 2x - 1}{6} \\\\\frac{12x - 15 - 2x - 1}{6}\\\\\frac{10x - 16}{6}\\\\ \frac{2(5x - 8)}{2 * 3} \\\\ \frac{5x - 8}{3}

4 0
3 years ago
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
A= 1/2 h (a+b) solve for h
Valentin [98]
"<span>A= 1/2 h (a+b) solve for h"
</span>
bh + 1/2h
7 0
3 years ago
Which system of equations has no solutions?
baherus [9]

Answer:

A

Step-by-step explanation:the lines do not cross therefor the two equations do not share a solution.

5 0
3 years ago
Which set of three numbers could be the side lengths of a triangle?
Genrish500 [490]
D.). I think. yeah because 2 side are equal and 1 isn't.
7 0
3 years ago
Read 2 more answers
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