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Minchanka [31]
3 years ago
6

nicole will rent a car for the weekend. she can choose one of two plans the first plan has an initial fee of $53 and cost an add

itional $0.08 per mile driven. the second plan has an initial fee of $48 and costs an additional $0.10 per mile driven.​
Mathematics
1 answer:
faltersainse [42]3 years ago
8 0

?

Step-by-step explanation:

what's the question??

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Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.
Temka [501]

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.

4 0
3 years ago
1. What is a stock? a. A bank account b. A type of bank c. Shares that are given to only rich people who own companies d. Shares
Sophie [7]

Answer:

D

Step-by-step explanation:

Stocks represent ownership in companies. Anyone can purchase publicly traded stock shares.

3 0
3 years ago
Please help me. I will give first person brainliest.
pychu [463]

Question 3:

Since, EG = EF + FG

64 = 2x-2+3x+1

64 = 5x -1

65 = 5x

x = 13

So, EF = 2x-2

= (2 \times 13)-2 = 26-2=24

So, EF = 24 units.

Question 4:

Since, XY \cong YZ

So, XY = YZ

9x-7 = 3x+5

6x = 12

So, x =2

Question 5:

Since, XY \cong YZ

By being congruent, the measures of the segments are equal.

So, 9x-7 = 3x+5

9x - 3x = 12

6x = 12

So, x = 2

Now, XZ = XY + YZ

XY = 9x - 7 = (9 \times 2) - 7 = 11

YZ = 3x + 5 = (3 \times 2) + 5 = 11

XZ = XY + YZ = 11 + 11 = 22 units.

Question 6:

It is given that m \angle ABC \cong m \angle DBE

It means that the measures of these two angles are equal.

Since, measure of angle ABC is 35 degrees.

Therefore, the measure of angle  DBE is 35 degrees.

7 0
4 years ago
The solutions to a quadratic equation are 6 and 2/3. Which quadratic function is related to this equation?
alekssr [168]

Answer:

The quadratic equation for the given roots  6 , \frac{2}{3} is y = 3 x² - 20 x + 12

Step-by-step explanation:

Given as :

The roots of the quadratic equation are 6 , \frac{2}{3}

Let The standard quadratic equation

a x² + b x + c = 0

So, the roots are \alpha = 6

And \beta =  \frac{2}{3}

Sum of roots = \alpha +\beta

i.e  \alpha +\beta  = 6 + \frac{2}{3} = \frac{20}{3}            

<u>The quadratic equation</u>

y = (x - α) (x - β)

Or, y = (x - 6) (x -  \frac{2}{3})

Or, y = x² -  \frac{2}{3} x - 6 x + (6 ×  \frac{2}{3})

Or, y =  x² - ( \frac{2}{3} + 6) x + 4

Or, y = x² - (\frac{2+18}{3} ) x + 4

Or, y =  x² - \frac{20}{3}x + 4

Or, y = 3 x² - 20 x + 12

Hence, The quadratic equation for the given roots  6 , \frac{2}{3} is y = 3 x² - 20 x + 12 . Answer

6 0
3 years ago
ASAP I WILL GIVE BRAINEST <br><br> GEOMETRY
Wewaii [24]
Pretty sure the answer is SAS because SSS only work on right triangles and this triangle has an angle
4 0
3 years ago
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