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3 years ago

Which is the endpoint of a ray? Point R Point S Point T Point U

Mathematics

2 answers:

slava [35]3 years ago

8 0

Point S is a ray!!

Hope this help!!

lapo4ka [179]3 years ago

5 0

9514 1404 393

Answer:

Point S

Step-by-step explanation:

A ray extends in one direction from its end point. Point S is the only point shown in the diagram that does not have parts of a line extending in opposite directions from it.

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Factor completely 3x2 − x − 4.

vova2212 [387]

If you would like to factor completely 3 * x^2 - x - 4, you can do this using the following steps:

<span>3 * x^2 - x - 4 = (3 * x - 4) * (x + 1) = (3x - 4) * (x + 1)
</span>
The correct result would be (3x - 4) * (x + 1<span>).</span>

3 0

3 years ago

A price of a motorcycle is Rs 175000. The year rate of deprecation is 4%. Find the price of motorcycle after 3 years

sasho [114]

$FV = PV (1 - \frac{r}{100})^n\\FV = 175000 (1 - \frac{4}{100})^3\\$

FV = Rs 154828.8

8 0

2 years ago

Evaluate the following expressions if ×=2,y=3 and Z=4. a.4×+×-3over ×+10 b.2y-5yover y+5. c.yz-×. d.9+y+zover y+2

sammy [17]

Answer:

A= 7 11/12 or 7.916 or 95.12

B= 4 1/8 or 4.125 or 33/8

C= 12 4/5 or 12.8 or 64/5

6 0

3 years ago

6.1.3 quiz pythagorean theorem question 9

Olegator [25]

Answer: 14.73

Step-by-step explanation:

The given triangle is a right angle triangle.

EF^2 + DF^2 = ED^2

The hypotenuse is |ED| while the two shorter legs are |EF| and |DF|.

We can then apply the Pythagoras Theorem to find the length of EF.

(EF)^2 + (DF)^2 = (ED)^2

(EF)^2 + (12)^2 = (19)^2

(EF)^2 + 144 = 361

(EF)^2 = 361 - 144

(EF)^2 = 217

EF = 14.73

6 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