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Radda [10]
3 years ago
14

The cost​ T, in hundreds of​ dollars, of tuition and fees at many community colleges can be approximated by Upper T equals one h

alf c plus 1
​, where c is the number of credits for which a student registers. Graph the equation and use the graph to estimate the cost of tuition and fees when a student registers for 2 three​-credit courses.
Mathematics
1 answer:
Rainbow [258]3 years ago
7 0
Course set of 8 and 2/3rds
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What is 84 ft. to 84 sq ft.
svetlana [45]
12 sq. in. in a foot i think 
5 0
3 years ago
Select the correct answer(click the photo)
MatroZZZ [7]

Answer:

B.

General Formulas and Concepts:

  • Order of Operations: BPEMDAS
  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define function</u>

f(x) = 8x + 4

<u>Step 2: Prep</u>

  1. Rewrite:                         y = 8x + 4
  2. Swap <em>x</em> and <em>y</em>:               x = 8y + 4

<u>Step 3: Find Inverse (Solve for </u><em><u>y</u></em><u>)</u>

  1. Subtract 4 on both sides:                    x - 4 = 8y
  2. Divide both sides by 8:                       (x - 4)/8 = y
  3. Rewrite:                                                 y = (x - 4)/8
  4. Rewrite:                                                f⁻¹(x) = (x - 4)/8
3 0
3 years ago
Solve this system of equations: y = x2 – 3x 12 y = –2x 14 1.isolate one variable in the system of equations, if needed. y = x2 –
Dahasolnce [82]

Answer:

our answer is X=2 or X= -1

4 0
2 years ago
calculate the unit rate of change for each linear function using the formula. (4,9) (8,12) (16,18) (32,30)
jenyasd209 [6]

Answer:

The unit rate of change is of \frac{3}{4}

Step-by-step explanation:

We are given a set of points (x,y).

To find the unit rate of change, we take two points, and divide the change in y by the change in x.

Points (4,9) and (8,12)

Change in y: 12 - 9 = 3

Change in x: 8 - 4 = 4

Unit rate of change: \frac{3}{4}

7 0
3 years ago
19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical
Tems11 [23]

EXPLANATION

Since we have the function:

f(x)=\frac{x^2-8x+15}{x^2}

Vertical asymptotes:

For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.

Taking the denominator and comparing to zero:

x+5=0

The following points are undefined:

x=-5

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1y=\frac{1}{1}\mathrm{The\:horizontal\:asymptote\:is:}y=1

In conclusion:

\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1

4 0
1 year ago
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