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2 years ago

PLS HELP ME ON THIS YOU CAN DO ALL THE QUESTIONS PLSSSS

Mathematics

1 answer:

Ivahew [28]2 years ago

5 0

Answer:

The first is $\frac{4}{10}$

The second is $\frac{10}{16}$

The third is $\frac{3}{10}$

The fourth is $\frac{6}{21}$

The fifth is $\frac{3}{5}$

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PLEASE HELP-

salantis [7]

Answer:

1 3/4 simplified or 1 12/16 not simplified

5 0

3 years ago

A graduated cylinder is filled with 36\pi36π36, pi cm^3 3 start superscript, 3, end superscript of liquid. The liquid is poured

lilavasa [31]

By definition, the volume of a cylinder is given by:
V = π * r ^ 2 * h
Where,
r: cylinder radius
h: height
Clearing h we have:
h = (V) / (π * r ^ 2)
Substituting values:
h = (36π) / (π * 3 ^ 2)
h = (36π) / (9π)
h = (36π) / (9π)
h = 4 cm
Answer:
The height of the liquid will be in the new cylinder about:
h = 4 cm

5 0

3 years ago

Read 2 more answers

o The day before the exam, Rebecca does a practice run. She attempts 20 glass vases and breaks 3 of them. What is her success ra

Rashid [163]

Answer:

If the goal is to break the vases: 3/20 = 15.0% or 0.15

If the goal is to not break the vases: 17/20 = 85.0% or 0.85

8 0

3 years ago

A company wishes to manufacture some boxes out of card. The boxes will have 6 sides (i.e. they covered at the top). They wish th

Serhud [2]

Answer:

The dimensions are, base $b=\sqrt[3]{200}$ , depth $d=\sqrt[3]{200}$ and height $h=\sqrt[3]{200}$ .

Step-by-step explanation:

First we have to understand the problem, we have a box of unknown dimensions (base $b$ , depth $d$ and height $h$ ), and we want to optimize the used material in the box. We know the volume $V$ we want, how we want to optimize the card used in the box we need to minimize the Area $A$ of the box.

The equations are then, for Volume

$V=200cm^3 = b.h.d$

For Area

$A=2.b.h+2.d.h+2.b.d$

From the Volume equation we clear the variable $b$ to get,

$b=\frac{200}{d.h}$

And we replace this value into the Area equation to get,

$A=2.(\frac{200}{d.h} ).h+2.d.h+2.(\frac{200}{d.h} ).d$

$A=2.(\frac{200}{d} )+2.d.h+2.(\frac{200}{h} )$

So, we have our function $f(x,y)=A(d,h)$ , which we have to minimize. We apply the first partial derivative and equalize to zero to know the optimum point of the function, getting

$\frac{\partial A}{\partial d} =-\frac{400}{d^2}+2h=0$

$\frac{\partial A}{\partial h} =-\frac{400}{h^2}+2d=0$

After solving the system of equations, we get that the optimum point value is $d=\sqrt[3]{200}$ and $h=\sqrt[3]{200}$ , replacing this values into the equation of variable $b$ we get $b=\sqrt[3]{200}$ .

Now, we have to check with the hessian matrix if the value is a minimum,

The hessian matrix is defined as,

$H=\left[\begin{array}{ccc}\frac{\partial^2 A}{\partial d^2} &\frac{\partial^2 A}{\partial d \partial h}\\\frac{\partial^2 A}{\partial h \partial d}&\frac{\partial^2 A}{\partial p^2}\end{array}\right]$

we know that,

$\frac{\partial^2 A}{\partial d^2}=\frac{\partial}{\partial d}(-\frac{400}{d^2}+2h )=\frac{800}{d^3}$

$\frac{\partial^2 A}{\partial h^2}=\frac{\partial}{\partial h}(-\frac{400}{h^2}+2d )=\frac{800}{h^3}$

$\frac{\partial^2 A}{\partial d \partial h}=\frac{\partial^2 A}{\partial h \partial d}=\frac{\partial}{\partial h}(-\frac{400}{d^2}+2h )=2$

Then, our matrix is

$H=\left[\begin{array}{ccc}4&2\\2&4\end{array}\right]$

Now, we found the eigenvalues of the matrix as follow

$det(H-\lambda I)=det(\left[\begin{array}{ccc}4-\lambda&2\\2&4-\lambda\end{array}\right] )=(4-\lambda)^2-4=0$

Solving for $\lambda$ , we get that the eigenvalues are: $\lambda_1=2$ and $\lambda_2=6$ , how both are positive the Hessian matrix is positive definite which means that the function $A(d,h)$ is minimum at that point.

4 0

3 years ago

Choose the equation that represents a line that passes through points (-6, 4) and (2.0).

Andrew [12]

58/627=916 I hope This help

8 0

3 years ago

Read 2 more answers