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
Zina [86]
3 years ago
9

Graph the parabola. y= - 5over2 X^2 -20x -35 plot the vertex and four additional points, two on each side of the vertex. Then cl

ick on the graph icon. I need the plots
Mathematics
1 answer:
stich3 [128]3 years ago
8 0
Wheres the graph ;/ or an link or image
You might be interested in
Wayne spends 1/6th of an hour walking his dog. How much time does Wayne spend walking his dog in a week? *
Mila [183]

Answer:

10 minutes

Step-by-step explanation:

This is somewhat of a trick question and also incomplete.

Assuming he walks his dog 1/6th of an hour each week, he walks his dog 10 minutes per week.

If this is incorrect please tell me the complete question

3 0
3 years ago
Solve the following equation -4y+8=4[2y-2]-
aksik [14]
The answer is y=4/3. Hope this helps.
4 0
3 years ago
If a and b are positive numbers, find the maximum value of f(x) = x^a(2 − x)^b on the interval 0 ≤ x ≤ 2.
Ad libitum [116K]

Answer:

The maximum value of f(x) occurs at:

\displaystyle x = \frac{2a}{a+b}

And is given by:

\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

Step-by-step explanation:

Answer:

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0

And we want to find the maximum value of f(x) on the interval [0, 2].

First, let's evaluate the endpoints of the interval:

\displaystyle f(0) = (0)^a(2-(0))^b = 0

And:

\displaystyle f(2) = (2)^a(2-(2))^b = 0

Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:

\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]

By the Product Rule:

\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}

Set the derivative equal to zero and solve for <em>x: </em>

\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]

By the Zero Product Property:

\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0

The solutions to the first equation are <em>x</em> = 0 and <em>x</em> = 2.

First, for the second equation, note that it is undefined when <em>x</em> = 0 and <em>x</em> = 2.

To solve for <em>x</em>, we can multiply both sides by the denominators.

\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))

Simplify:

\displaystyle a(2-x) - b(x) = 0

And solve for <em>x: </em>

\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\  \frac{2a}{a+b} &= x  \end{aligned}

So, our critical points are:

\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b

This can be simplified to:

\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

Since <em>a</em> and <em>b</em> > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.

To confirm that this is indeed a maximum, we can select values to test. Let <em>a</em> = 2 and <em>b</em> = 3. Then:

\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)

The critical point will be at:

\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8

Testing <em>x</em> = 0.5 and <em>x</em> = 1 yields that:

\displaystyle f'(0.5) >0\text{ and } f'(1)

Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.

Therefore, the maximum value of f(x) occurs at:

\displaystyle x = \frac{2a}{a+b}

And is given by:

\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

5 0
3 years ago
Is anyone good a maths?I really need your help for a test coming up?​
tester [92]

Answer:

i am, but im not good at high school,

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What number can you put in the missing spot of the domain to keep the table a function? Select all that apply.​
Bingel [31]

Answer:

1,8 and 5,10

Step-by-step explanation:

5 0
3 years ago
Other questions:
  • Eight less than four times the number c is twenty.<br> What is the value of c?
    10·2 answers
  • Which example of artificial selection is caused indirectly by human activity?
    14·1 answer
  • I need to know how to work<br> This
    10·1 answer
  • (5/3)^-2<br> (9/11)^-1 <br> brainlest
    7·2 answers
  • Sara got 88, 90, 75, 78, 100, 54, 90, 100, 45, and 60 on her English tests
    5·2 answers
  • After John worked at a job for 10 years, his salary doubled. If he started at $x, hls salary after 10 years is​
    14·1 answer
  • In a probability experiment Eric flipped a coin 30 times the coin landed on heads 20 times what is the ratio of heads to tails i
    5·2 answers
  • ⅕ + ⅕ + ⅕ +⅕ <br><br> 6 x ⅕ <br><br> Which is GREATER?
    10·1 answer
  • Find the midpoint of the line segment joining the points (7,1) and (-1,-1)
    8·1 answer
  • 2. Write an algebraic equation and solve it
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!