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3 years ago

Why do banks charge fees?

Mathematics

1 answer:

Ivahew [28]3 years ago

3 0

They charge fees so they can cover there costs and return value to shareholders

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Please help 10 points !!

enyata [817]

The will intersect at a point with a positive x coordinate.

We can tell this because they actually already have a shared space on the graph. If you plot the 3 points of g(x) given, you'll see that f(x) and g(x) both share the coordinate (1, 3). As a result, we know that they do intersect and it is where x (1) is a positive number.

4 0

3 years ago

Match the correct graph to the story.

emmainna [20.7K]

Answer:

The graph for number 1 (about the bunnies) is C.

The graph for number 2 (about the kettle) is A.

Step-by-step explanation:

C because it is an exponential function.

And A because it is the only one decreasing in temperature.

3 0

3 years ago

Ron has a bag containing 3 green pears and 4 red pears. He randomly selects a pear then randomly selects another pear, without r

oksian1 [2.3K]

The files did not load correctly

4 0

3 years ago

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

(-13, 3) (5, 12) (-13, 0) Is this relation a function?

Natasha2012 [34]

Answer:

No.

Step-by-step explanation:

No because it is a one-to-many relation .

-13 maps to 3 and to 0.

6 0

3 years ago