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
Alenkinab [10]
3 years ago
12

I REALLY NEED HELP ASAP

Mathematics
1 answer:
guapka [62]3 years ago
6 0

The Correct option is -4

because every other terms result in value = 4, where's -4 is the only different one hence doesn't belongs yo other three

You might be interested in
Lola saved $86.25 in 3 months. She saved $30.75 in the first month and $7.99 in the second month. How much money did Lola save i
pickupchik [31]
The correct answer is 47 dollars and 51 cents.
5 0
3 years ago
Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |
dsp73

Looks like f(x,y)=x^2+y^2-x^2y+7.

f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1

f_y=2y-x^2=0\implies2y=x^2

  • If x=0, then y=0 - critical point at (0, 0).
  • If y=1, then x=\pm\sqrt2 - two critical points at (-\sqrt2,1) and (\sqrt2,1)

The latter two critical points occur outside of D since |\pm\sqrt2|>1 so we ignore those points.

The Hessian matrix for this function is

H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}

The value of its determinant at (0, 0) is \det H(0,0)=4>0, which means a minimum occurs at the point, and we have f(0,0)=7.

Now consider each boundary:

  • If x=1, then

f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4

which has 3 extreme values over the interval -1\le y\le1 of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

  • If x=-1, then

f(-1,y)=8-y+y^2

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

  • If y=1, then

f(x,1)=8

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

  • If y=-1, then

f(x,-1)=2x^2+8

which has 3 extreme values, but the previous cases already include them.

Hence f(x,y) has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0
3 years ago
If x+1/x= 3 then x-1/x = ?
Alekssandra [29.7K]

Answer:

If x+1/x= 3 Then the answer for this equation (x-1/x=? ) will be...

Step-by-step explanation:

5 0
3 years ago
In the XY coordinate plane, the slope of a line is 4 and it passes through the point (1,3). Find the equation of a line that is
Sergeeva-Olga [200]

Answer:

E?

Step-by-step explanation:

I'm sorry of this answer is wrong

7 0
3 years ago
find two consecutive odd intergers that twice the larger is fifteen more than three times the smaller
Svetach [21]

Answer:

11, 13

Step-by-step explanation:

an odd number can be represented by 2n+1

since they are consecutive, the larger odd number will be 2n + 1 + 2

now,

2(2n+3) = 2n + 1 +15

solving this eqn, we get n = 5

so the two numbers are (2n+1) = 11 and 11+2= 13

4 0
3 years ago
Read 2 more answers
Other questions:
  • 2/7, solve for n, explain too please
    14·2 answers
  • What is the equation of the line through (4,5) and (0,-3)
    15·2 answers
  • <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3-a%7D%7B6%7D%20-%5Cfrac%7B6-a%7D%7B3%7D%20%3Da%2B1%3Cbr%3E%0A%20" id="TexFormula
    15·1 answer
  • At a lumber company, shelves are sold in 4 types of wood, 3 different widths and 3 different lengths. How many different types o
    11·1 answer
  • Clara lends half her collection of formal attires to her sister, Susan. Clara then buys four more attires. If she has 12 attires
    9·1 answer
  • Can someone please help 20 points!!
    6·2 answers
  • Which expression is equivalent to this quotient?
    13·1 answer
  • 3/5+1/4=<br> pls help !!!!!
    12·1 answer
  • Which scatter plot shows the relationship
    15·1 answer
  • Simplify: j + 5 + 5c
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!