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
Contact [7]
3 years ago
14

Is this relation a function? Justify your answer.

Mathematics
1 answer:
kap26 [50]3 years ago
8 0

Answer:

it is only in the positive so i would say d because they all only have one y-value each and they are all positive though so i am saying D.

You might be interested in
Any help plzzz I’ll appreciate it so much thank u
docker41 [41]

9514 1404 393

Answer:

  5√5

Step-by-step explanation:

Use the distance formula:

  d = √((x2 -x1)² +(y2 -y1)²)

  d = √((-7-(-2))² +(-7-3)²) = √((-5)² +(-10)²) = √(25 +100)

  d = √125 = √(25·5)

  d = 5√5 . . . . distance between the points

3 0
3 years ago
Directions: Find the inverse of each number.
Alisiya [41]

Answer:

Step-by-step explanation:

Additive inverse:

-3=3

14=-14

-28=28

-1/5=1/5

Multiplicative Inverse:

2/3=3/2

6/18=18/6

5=1/5

11=1/11

3 0
2 years ago
Read 2 more answers
Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d
aliya0001 [1]

The Lagrangian

L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)

has critical points where the first derivatives vanish:

L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}

L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}

L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}

L_\lambda=x^4+y^4+z^4-13=0

We can't have x=y=z=0, since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of \pm\sqrt[4]{13}. For example, if y=z=0, then x^4=13\implies x=\pm\sqrt[4]{13}.

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find \lambda=-\frac1{\sqrt{26}} (taking the negative root because x^2,y^2,z^2 must be non-negative), and we can immediately find the critical points from there. For example, if z=0, then x^4+y^4=13. If both x,y are non-zero, then x^2=y^2=-\frac1{2\lambda}, and

xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}

\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}

and for either choice of x, we can independently choose from y=\pm\sqrt[4]{\frac{13}2}.

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then x^2=y^2=z^2=-\frac1{2\lambda}. We have

xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}

\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}

and similary y,z have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate f at each critical point; you should end up with a maximum value of \sqrt{39} and a minimum value of \sqrt{13} (both occurring at various critical points).

Here's a comprehensive list of all the critical points we found:

(\sqrt[4]{13},0,0)

(-\sqrt[4]{13},0,0)

(0,\sqrt[4]{13},0)

(0,-\sqrt[4]{13},0)

(0,0,\sqrt[4]{13})

(0,0,-\sqrt[4]{13})

\left(\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)

\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

5 0
3 years ago
Answer number 5 click image.
natita [175]
That makes 55200*0.62=34224 people.
So 34224 people voted.
3 0
4 years ago
Read 2 more answers
Circle theorem - easy work
makvit [3.9K]
Angle in a semi circle is 90°
<a = 180-90-<b = 90-28 = 62°

opposite angles of a cyclic quadrilateral sums up to 180°
<b + <a = 180
<b = 180-62
<b = 118°
8 0
4 years ago
Other questions:
  • What does this complex fraction equal??? Please help me
    8·1 answer
  • Jordan places two boards end to end to make one shelf. The first board is 47-100 meter long.The second board is 5-10 meter long.
    14·1 answer
  • I really need help!
    6·1 answer
  • Hiram raises earthworms. In a square of compost 4 ft by 4 ft, he can have 1000 earthworms. How many earthworms can he have if hi
    10·2 answers
  • What is the surface area of this solid
    10·1 answer
  • X/3 + 5 = 6.8<br><br> The x/3 is a fraction! please I need an answer!!!
    5·1 answer
  • Need these in 10 mins: Solve by substitutions<br> 1) X+Y=6<br> 3x+2y=15<br> 2 2x+Y=20<br> 3) 6x-SY=r
    9·1 answer
  • Imal ikoga iz Bosne?​
    15·2 answers
  • Translate the sentence into an equation
    5·2 answers
  • Can someone please help me solve this…
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!