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3 years ago

When a scatterplot is created from a table of values, which statement is correct

Mathematics

2 answers:

olga nikolaevna [1]3 years ago

8 0

Answer:

Step-by-step explanation:

lorasvet [3.4K]3 years ago

7 0

<span>It is possible for two points to have the same x-coordinate and the same y-coordinate. </span>

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Which constant could each equation be multiplied by to eliminate the x variable in this system of equations? 4x + 5y = 62 5x − 3

qwelly [4]

Answer:

Multiply equation $4x+5y=62$ by 5 and $5x-3y=22$ by 4

Step-by-step explanation:

Given:

Equations are $4x+5y=62,5x-3y=22$

To find: constant by which each equation should be multiplied to eliminate the variable $x$ in the given system of equations

Solution:

$4x+5y=62...(i)\\5x-3y=22...(ii)$

Multiply equation (i) by 5 and (ii) by 4 to get the following equations:

$20x+25y=310\\20x-12y=88$

On subtracting these equations, we get

$(20x+25y)-(20x-12y)=310-88\\\\13y=222$

So, the variable $x$ gets eliminated

7 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.