1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
marusya05 [52]
2 years ago
8

Shown here is the graph of a function f.

Mathematics
1 answer:
vazorg [7]2 years ago
6 0

Answer:

C) They are the same.

Step-by-step explanation:

<u>The function f(x) has points as per graph:</u>

  • f(1) = 9
  • f(3) = 1

<u>So the points are:</u>

  • (1, 9) and (3, 1)

<u>Rate of change for f(x):</u>

  • (1-9)/(3-1) = - 8/2 = -4

<u>The function g(x) has points:</u>

  • g(1) = -1² + 10 = 9
  • g(3) = -3² + 10 = 1

<u>So the points are:</u>

  • (1, 9) and (3, 1)

<u>Rate of change for f(x):</u>

  • (1-9)/(3-1) = - 8/2 = -4

As we see both functions have same rate of change over the interval [1, 3]

<u>The answer choice is: </u>C) They are the same

You might be interested in
Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is
strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is 121y''+110y'-24y=0. We propose that the solution of this equations is of the form y = Ae^{rt}. Then, by replacing the derivatives we get the following

121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

121r^2+110r-24=0

Recall that the roots of a polynomial of the form ax^2+bx+c are given by the formula

x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

r_1 = -\frac{12}{11}

r_2 = \frac{2}{11}

So, in this case, the general solution is y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

c_1 + c_2 = 1

c_1\frac{-12}{11} + c_2\frac{2}{11} = 0(or equivalently c_2 = 6c_1

By replacing the second equation in the first one, we get 7c_1 = 1 which implies that c_1 = \frac{1}{7}, c_2 = \frac{6}{7}.

So y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}

b) By using y(0) =0 and y'(0)=1 we get the equations

c_1+c_2 =0

c_1\frac{-12}{11} + c_2\frac{2}{11} = 1(or equivalently -12c_1+2c_2 = 11

By solving this system, the solution is c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}

Then y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2

By plugging the values of y_1 and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

e^{\int -p(x) dx}

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}

Note that this function is always positive, and thus, never zero. So y_1, y_2 is a fundamental set of solutions.

8 0
2 years ago
A business owes $8000 on a loan. Every month, the business pays 12 of the amount remaining on the loan.
steposvetlana [31]
<span>$125 is to be paid during the 6th month.</span>
6 0
3 years ago
100 POINTS IF U GET THIS RIGHT!
iVinArrow [24]

Answer:

That is the answer

Hope this helps!

3 0
2 years ago
Read 2 more answers
Find the equation of the line through point (2,2) and parallel to y=x+4. Use a forward slash (i.e.”/“) for fractions (e.g. 1/2 f
aleksandr82 [10.1K]

Answer:

The equation of the line is, y = x

Step-by-step explanation:

The constraints of the required linear equation are;

The point through which the line passes = (2, 2)

The line to which the required line is parallel = y = x + 4

Two lines are parallel if they have the same slope, therefore, we have;

The slope of the line, y = x + 4 is m = 1

Therefore, the slope of the required line = 1

The equation of the required lime in point and slope form becomes;

y - 2 = 1 × (x - 2)

∴ y = x - 2 + 2 = x

The equation of the required line is therefore, y = x

5 0
3 years ago
Multiply (-8x - 9)(-9x2 + 7x - 9).
Misha Larkins [42]
= <span>(-8x - 9)(-9x2 + 7x - 9)

= 72x</span>³ - 56x² + 72x + 81x² - 63x + 81

= 72x³ + 25x² + 9x + 81

In short, Your Answer would be: Option D

Hope this helps!
3 0
3 years ago
Read 2 more answers
Other questions:
  • I’m confused please help and thanks
    6·1 answer
  • What is the circumference of a circle whose area is 100<img src="https://tex.z-dn.net/?f=%20%5Cpi%20" id="TexFormula1" title=" \
    14·1 answer
  • Answer dis in a big hurry and will give brainliest!!!
    12·1 answer
  • Can someone help me with these 2 questions?
    6·1 answer
  • Lori wants to put a tarp over their family's circular pool to keep the leaves out. The pool has a diameter of 20 feet. How many
    11·1 answer
  • Help answer pleaseeeeeee
    7·1 answer
  • Length = 4<br> Width = ?<br> Area = 60
    14·1 answer
  • HALLPPP its worth 50 points
    5·1 answer
  • What is 1800/1254<br><img src="https://tex.z-dn.net/?f=1800%20%5Cdiv%201254" id="TexFormula1" title="1800 \div 1254" alt="1800 \
    10·1 answer
  • The line passes through the point (3,0) and has a slope of -3
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!