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tangare [24]
4 years ago
7

You have five different varieties of flowers and you want to put 3 different varieties in each vase how many cases will you make

without duplicating a vase
Mathematics
1 answer:
Marat540 [252]4 years ago
4 0

Answer:

10

Step-by-step explanation:

Given:

There are 5 different varieties of flowers.

To place 3 different varies in each vase.

To find how many ways we can do so without duplicating a vase.

Solution:

We need to choose 3 different flowers out of 5 different variety of flowers.

In order to do so we use combination.

We can place the flowers in 5C3 ways.

NCr=\frac{N!}{(n-r)!(r!)}

5C3=\frac{5!}{(5-3)!3!} = \frac{5\times 4\times 3\times2\times1}{(2\times1)(3\times2\times1)}=\frac{5\times4}{2\times1}=\frac{20}{2}=10

Thus, we can make vase in 10 different ways without duplicating a vase.

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siniylev [52]

For this case we have:

By definition, two points are needed to find the equation of a line. We have a series of points given by:

(x, y) = (time, cost)

We chose two points:

(t1, c1) = (1,180)\\(t2, c2) = (2,240)

By definition, the formula of the slope is given by:

m =\frac{(c2-c1)}{(t2-t1)}

Substituting:

m =\frac{240-180}{2-1}

m =\frac{60}{1}\\m = 60

Being a line of the form y = ax + b, we substitute a point and the slope found to find the cut point b.

180 = 60 * 1 + b\\180 = 60 + b\\b = 180-60\\b = 120

Thus, the equation of the line is given by:

y = 60x + 120

Answer:

Option C

6 0
4 years ago
Read 2 more answers
Check the true statements below:
valentinak56 [21]

Answer:

a) False

b) False

c) True

d) False

e) False

Step-by-step explanation:

a. A single vector by itself is linearly dependent. False

If v = 0 then the only scalar c such that cv = 0 is c = 0. Hence, 1vl is linearly independent. A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, only a single zero vector is linearly dependent, while any set consisting of a single nonzero vector is linearly independent.

b. If H= Span{b1,....bp}, then {b1,...bp} is a basis for H. False

A sets forms a basis for vector space, only if it is linearly independent and spans the space. The fact that it is a spanning set alone is not sufficient enough to form a basis.

c. The columns of an invertible n × n matrix form a basis for Rⁿ. True

If a matrix is invertible, then its columns are linearly independent and every row has a pivot element. The columns, can therefore, form a basis for Rⁿ.

d.  In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix. False

Row operations can not affect linear dependence among the columns of a matrix.

e. A basis is a spanning set that is as large as possible. False

A basis is not a large spanning set. A basis is the smallest spanning set.

3 0
3 years ago
Find the slope of the line that passes through (9, 6) and (5, 5).
Alecsey [184]

Answer:

1/4

Step-by-step explanation:

y2-y1/x2-x1=m

6-5/9-5=1/4

7 0
3 years ago
What are the first three terms of the sequence defined by the recursive function an = an-1-(an-2-4), when a5 =-2 and a6 = 0?
Wittaler [7]

Answer:

<em>6, 10 and 8</em>

Step-by-step explanation:

Given the recursive function an = an-1-(an-2 - 4), when a5 =-2 and a6 = 0?

a6 = a5 - (a4 - 4)

0 = -2  - (14 - 4)

2 = - (a4 - 4)

-2 = a4 - 4

a4 = -2 + 4

a4 = 2

a5 = a4 - (a3 - 4)

-2 = 2  - (a3 - 4)

-2-2 = - (a3 - 4)

-4 = -(a3 - 4)

4 = a3 - 4

a3 = 4+4

a3 = 8

Also

a4 = a3 - (a2 - 4)

2 = 8 - (a2 - 4)

2-8 = - (a2 - 4)

-6 = -(a2 - 4)

6 = a2 - 4

a2 = 6+4

a2 = 10

a3 = a2 - (a1 - 4)

8 = 10 - (a1 - 4)

8-10 = - (a1 - 4)

-2 = -(a1 - 4)

2 = a1 - 4

a1 = 2+4

a1 = 6

<em>Hence the first 3 terms are 6, 10 and 8</em>

6 0
3 years ago
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ivolga24 [154]
Ones because it is the last place value
5 0
3 years ago
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