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UNO [17]
2 years ago
5

When creating a histogram, the first recorded scale is 0 to 10. What should the second scale indicate?

Mathematics
1 answer:
klio [65]2 years ago
6 0

Answer:

A) 10-20

Step-by-step explanation:

Its A because it started with 0 and it went to 10 so it adds 10 each time cause it gave us the hint with 0 to 10,

I hope this helps

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2 years ago
The cost of the tank and filter system is $280. The base of the tank is a square with a perimeter of 60 inches what is the side
Kaylis [27]

Answer:

The side length of of the base of the tank is 15 inches.

Step-by-step explanation:

We are given the following in the question:

Cost of the tank and filter system = $280

Perimeter of tank base = 60 inches

The perimeter of tank is in the shape of square.

We have to find the side length of the base of the tank.

Perimeter of tank base = Perimeter of square =

P = 4a\\\text{where a is the side of square}\\60 = 4a\\\Rightarrow a  = 15\text{ inches}

Thus, the side length of of the base of the tank is 15 inches.

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
The inverse of Fx) is a function.<br> FOD<br> A. True<br> B. False
anastassius [24]
Not all functions have inverse functions. Those that do are called invertible. For a function f: X → Y to have an inverse, it must have the property that for every y in Y, there is exactly one x in X such that f(x) = y.
8 0
3 years ago
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