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Ivanshal [37]
2 years ago
14

Jessica is deciding on her schedule for next semester. She must take each of the following classes: English 101, Spanish 102, Bi

ology 102, and College Algebra. If there are 15 sections of English 101, 9 sections of Spanish 102, 11 sections of Biology 102, and 15 sections of College Algebra, how many different possible schedules are there for Jessica to choose from? Assume there are no time conflicts between the different classes.
Mathematics
1 answer:
user100 [1]2 years ago
8 0
Just a random guess 22,275 tell me if it right
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Macy’s is having a sale in which all jeans at $30. The sales tax is 8%. If John buys n pairs of jeans during this sale, the tota
Amiraneli [1.4K]

This expression is correct. It shows the price John pays for the jeans on the left side of the + symbol (30n), and the amount of tax he has to pay on the right side of the symbol (0.08(30n)).

Hope this helps!

3 0
2 years ago
Match the features of the graph of the rational function.
Sunny_sXe [5.5K]

After applying <em>algebraic</em> analysis we find the <em>right</em> choices for each case, all of which cannot be presented herein due to <em>length</em> restrictions. Please read explanation below.

<h3>How to analyze rational functions</h3>

In this problem we have a rational function, whose features can be inferred by algebraic handling:

Holes - x-values that do not belong to the domain of the <em>rational</em> function:

x³ + 8 · x² - 9 · x = 0

x · (x² + 8 · x - 9) = 0

x · (x + 9) · (x - 1) = 0

x = 0 ∨ x = - 9 ∨ x = 1

But one root is an evitable discontinuity as:

y = (9 · x² + 81 · x)/(x³ + 8 · x² - 9 · x)

y = (9 · x + 81)/(x² + 8 · x - 9)

Thus, there are only two holes. (x = - 9 ∨ x = 1) Besides, there is no hole where the y-intercept should be.

Vertical asymptotes - There is a <em>vertical</em> asymptote where a hole exists. Hence, the function has two vertical asymptotes.

Horizontal asymptotes - <em>Horizontal</em> asymptote exists and represents the <em>end</em> behavior of the function if and only if the grade of the numerator is not greater than the grade of the denominator. If possible, this assymptote is found by this limit:

y = \lim_{x \to \pm \infty} \frac {9\cdot x + 81}{x^{2}+8\cdot x - 9}

y = 0

The function has a horizontal asymptote.

x-Intercept - There is an x-intercept for all x-value such that numerator is equal to zero:

9 · x + 81 = 0

x = - 9

There is a x-intercept.

Lastly, we have the following conclusions:

  1. How many holes? 2
  2. One <em>horizontal</em> asymptote along the line where y always equals what number: 0
  3. This function has x-intercepts? True
  4. One <em>vertical</em> asymptote along the line where x always equals what number: 1
  5. There is a hole where the y-intercept should be? False

To learn more on rational functions: brainly.com/question/27914791

#SPJ1

5 0
2 years ago
Find the solutions to the equation below.
taurus [48]

2x²+7x+3=0

2x²+(1+6)x+3=0

2x²+1x+6x+3=0

x(2x+1)+3(2x+1)=0

(2x+1)(x+3)=0

2x+1=0. x+3=0

2x=-1. x=-3

x=-1/2

7 0
3 years ago
Alan has read ​10% of a book. He has 81 more pages to finish. How many pages are there in the​ book?
rewona [7]

Answer:

90

Step-by-step explanation:

100%-10%=90%

90%->81

10%->9

100%->9+81=90

6 0
2 years ago
Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh
fredd [130]

With

\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k

we have

\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt

The vector field evaluated over this parameterization is

\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k

so the line integral is

\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt

=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4

6 0
3 years ago
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