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chubhunter [2.5K]
2 years ago
12

7. Which set of ordered pairs represents a function?

Mathematics
1 answer:
Likurg_2 [28]2 years ago
5 0

Answer:

a

Step-by-step explanation:

to be a function, every x has to have exactly one y, so there can't be the same number x twice wirh different y's

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Write the equation of a circle whose center is at (3,-2) and has a radius of 11.<br>Show your work!
Viefleur [7K]

Answer:

(x - 3)^{2} + (y + 2)^{2} = 121

Step-by-step explanation:

The equation of a circle has the following format:

(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}

In which r is the radius and the centre is the point (x_{0}, y_{0})

In this question:

Center at (3,-2), so x_{0} = 3, y_{0} = -2

Radius 11, so r = 11

Then

(x - 3)^{2} + (y - (-2))^{2} = 11^{2}

(x - 3)^{2} + (y + 2)^{2} = 121

3 0
3 years ago
Use cubic regression to find a function that fits the following points.
8090 [49]

Answer:

y = 2x³ + x² - 4x + 6

Step-by-step explanation:

General formula:

y = ax³ + bx² + cx + d

If you replace point (0, 6) in the general formula, you get:

6 = a(0)³ + b(0)² + c(0) + d

6 = d

If you replace points (-1, 9), (1, 5) and (2, 18) in the general formula, you get the following system of equations:

9 = a(-1)³ + b(-1)² + c(-1) + 6

9 = -a + b - c + 6     (eq. 1)

5 = a(1)³ + b(1)² + c(1) + 6

5 = a + b + c + 6     (eq. 2)

18 = a(2)³ + b(2)² + c(2) + d

18 = 8a + 4b + 2c + 6     (eq. 3)

Adding equation 1 to equation 2:

14 = 2b + 12

(14 - 12)/2 = b

b = 1

Multiplying equation 1 by 2:

18 = -2a + 2b - 2c + 12     (eq. 4)

Adding equation 4 to equation 3, and replacing with b value:

36 = 6a + 6b + 18

36 = 6a + 6 + 18

(36 - 6 - 18)/6 = a

a = 2

Replacing a and b values in equation 1:

9 = -2 + 1 - c + 6

c = -2 + 1 + 6  - 9

c = -4

8 0
3 years ago
Read 2 more answers
What is 2×2×2×5÷5÷100×100-100+1000​
ivanzaharov [21]

Answer:

908

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Use lagrange multipliers to find the point on the plane x â 2y + 3z = 6 that is closest to the point (0, 2, 4).
Arisa [49]
The distance between a point (x,y,z) on the given plane and the point (0, 2, 4) is

\sqrt{f(x,y,z)}=\sqrt{x^2+(y-2)^2+(z-4)^2}

but since \sqrt{f(x,y,z)} and f(x,y,z) share critical points, we can instead consider the problem of optimizing f(x,y,z) subject to x-2y+3z=6.

The Lagrangian is

L(x,y,z,\lambda)=x^2+(y-2)^2+(z-4)^2+\lambda(x-2y+3z-6)

with partial derivatives (set equal to 0)

L_x=2x+\lambda=0\implies x=-\dfrac\lambda2
L_y=2(y-2)-2\lambda=0\implies y=2+\lambda
L_z=2(z-4)+3\lambda=0\implies z=4-\dfrac{3\lambda}2
L_\lambda=x-2y+3z-6=0\implies x-2y+3z=6

Solve for \lambda:

x-2y+3z=-\dfrac\lambda2-2(2+\lambda)+3\left(4-\dfrac{3\lambda}2\right)=6
\implies2=7\lambda\implies\lambda=\dfrac27

which gives the critical point

x=-\dfrac17,y=\dfrac{16}7,z=\dfrac{25}7

We can confirm that this is a minimum by checking the Hessian matrix of f(x,y,z):

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

\mathbf H is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.

At this point, we get a distance from (0, 2, 4) of

\sqrt{f\left(-\dfrac17,\dfrac{16}7,\dfrac{25}7\right)}=\sqrt{\dfrac27}
8 0
3 years ago
A very slow man walks<br><br> 20 cm.in 0.5 left, what his<br><br> velocity in meter per second?
Gre4nikov [31]
Displacement(d)=0.2m
Time taken(t)= 30 s(1min=60 s)
Velocity (v)=?
We know that,
Velocity (v) = d/t
=0.0067
8 0
2 years ago
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