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LUCKY_DIMON [66]
2 years ago
15

Express in the form n : 1 . Give n as a decimal. 9 : 6

Mathematics
1 answer:
Serggg [28]2 years ago
3 0

Answer:

1.5 : 1

Step-by-step explanation:

N : 1 = 9 : 6

you can take 6 and divide by 6 to get 1. so do the same with 9. to get 9/6 = n

n = 1.5

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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
56/(7-9)^3-25/23-5*4
meriva
I got -28.08695652. It doesn't repeat nor does it terminate
4 0
2 years ago
Solve and state any extraneous solutions
Mice21 [21]

\frac{3x}{x - 1}  + 2 =  \frac{3}{x - 1}

\frac{3x}{x - 1} +  \frac{2(x - 1) }{ x - 1}   =  \frac{3}{x - 1}

\frac{3x}{x - 1}  +  \frac{2x - 2}{ x - 1}  =  \frac{3}{x - 1}

\frac{3x + 2x - 2}{ x - 1}  =  \frac{3}{x - 1}

3x + 2x - 2 = 3

5x - 2 = 3

5x = 3 + 2

5x = 5

x =  \frac{5}{5}

x = 1

3 0
3 years ago
What is the boundedness
DedPeter [7]

Answer:

Boundedness requires that there is only a bounded number of different kinds of parts

Step-by-step explanation:

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=4x%20%2B%203y%20%3D%200%20%5C%5C5y%20%2B%2053%20%3D%2011x%20%5C%5C%20" id="TexFormula1" title=
Anvisha [2.4K]

Answer:

x = 3, y = -4

Step-by-step explanation by substitution:

Solve the following system:

{4 x + 3 y = 0 | (equation 1)

5 y + 53 = 11 x | (equation 2)

Express the system in standard form:

{4 x + 3 y = 0 | (equation 1)

-(11 x) + 5 y = -53 | (equation 2)

Swap equation 1 with equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

4 x + 3 y = 0 | (equation 2)

Add 4/11 × (equation 1) to equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+(53 y)/11 = -212/11 | (equation 2)

Multiply equation 2 by 11/53:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+y = -4 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{-(11 x)+0 y = -33 | (equation 1)

0 x+y = -4 | (equation 2)

Divide equation 1 by -11:

{x+0 y = 3 | (equation 1)

0 x+y = -4 | (equation 2)

Collect results:

Answer: {x = 3 , y = -4

_____________________________________

Solve the following system:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for x:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Isolate terms with x to the left hand side.

Subtract 3 y from both sides:

{4 x = -3 y

5 y + 53 = 11 x

Hint: | Solve for x.

Divide both sides by 4:

{x = -(3 y)/4

5 y + 53 = 11 x

Hint: | Perform a substitution.

Substitute x = -(3 y)/4 into the second equation:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for y:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Isolate y to the left hand side.

Subtract 53 - (33 y)/4 from both sides:

{x = -(3 y)/4

(53 y)/4 = -53

Hint: | Solve for y.

Multiply both sides by 4/53:

{x = -(3 y)/4

y = -4

Hint: | Perform a back substitution.

Substitute y = -4 into the first equation:

Answer: {x = 3 , y = -4

7 0
2 years ago
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