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3 years ago

5.1.2 Exam: Semester 2 Exam

Mathematics

1 answer:

Nikolay [14]3 years ago

8 0

Answer:

See attached picture to view the graph

Step-by-step explanation:

Start by analyzing how this average idea works:

If only one member goes to the trip, it will cost him/her $1000+$200 = $1200.

If two members go to the trip, then they will share the cost as per the following: ($1000+ $200 + $200 = $1400) which they will be dividing into two people, thus costing each of them $700.

Notice that the general function that represents such average will be given by: $f(x) = \frac{1000+200x}{x}$

Plot such function in the two dimensional plane, and you will get the asymptotic behavior shown in the attached image.

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Write a cosine function that has a midline of 2, an amplitude of 4 and a period of pi/2.

exis [7]

y = Acos(Bx) + D;

D = 4, A = 2. Now T = 2π/B = 5π/8, B = 2π/(5π/8) = 16/5

WE get y = 2cos(16/5x) + 4

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2 years ago

Click to add text<br> Combine Like Terms:<br> 6y + 4z + 9x + 3y

boyakko [2]

Answer:

9y + 4z + 9x

Step-by-step explanation:

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3 years ago

Read 2 more answers

Points A, B, and C are collinear. Find the value of x given AB = 3x, BC = 5x + 10. and AC = 74.

Mars2501 [29]

Answer:

x = 8

Step-by-step explanation:

since it's a collinerar

or, AB+BC=AC

or, 3x+5x+10 = 74

or, 8x = 74-10

or, 8x = 64

or, x = 64/8

x = 8

6 0

4 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

3 years ago

3x = 15A) x= 4B) x= 5C) x = 6D) x= 12

xz_007 [3.2K]

3x = 15

Divide both-side of the equation by 3

3x/3 = 15/3

x=5

B) is the correct option

3 0

1 year ago