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Mazyrski [523]
3 years ago
10

The number of sweets s that each friend gets is the number of sweets in a box n divided by the number of friends f the sweets ar

e shared between. Enter a formula for the number of sweets each friend gets and enter the number of sweets each friend gets if there are 60 sweets and 12 friends.
Mathematics
1 answer:
Ainat [17]3 years ago
8 0

Answer:

The number of sweets each friend gets = 5

Step-by-step explanation:

The number of sweets s that each friend gets = the number of sweets in a box n ÷ number of friends f the sweets are shared between

s = n ÷ f

s = n/ f

if there are 60 sweets and 12 friends.

That is,

n = 60 sweets

f = 12 friends

Find s,

s = n/ f

= 60 / 12

= 5

s = 5 sweets

The number of sweets each friend gets = 5

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Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix  ⇒

\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

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2 years ago
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Hope this helps!

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S_A_V [24]

Answer:

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Step-by-step explanation:

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