All categories

2 years ago

Arrange the expressions in increasing order of their values.

Mathematics

1 answer:

Arada [10]2 years ago

3 0

Answer:

10^0*10^1-1,

10^0*10^1*1^10,

10^0+10^1*1^10,10^0+10^1+1^10

Step-by-step explanation:

You might be interested in

20 POINTS I need your help

disa [49]

The answer to your question is 405

6 0

3 years ago

Read 2 more answers

Consider the following equations x=4/t and y = 4-t^2 a. Solve the equation for y (eliminate the parameter.)

Nookie1986 [14]

Answer:

y=-16^a/x^2 + 4

Step-by-step explanation:

3 0

3 years ago

Johnson Chemicals is considering two options for its supplier portfolio. Option 1 uses two local suppliers. Each has a "unique-e

sashaice [31]

Answer:

a) P=0.0175

b) P=0.0189

Step-by-step explanation:

For both options we have to take into account that not only the chance of a "superevent" will disable both suppliers.

The other situation that will disable both is that both suppliers have their "unique-event" at the same time.

As they are, by definition, two independent events, we can calculate the probability of having both events at the same time as the product of both individual probabilities.

a) Then, the probability that both suppliers will be disrupted using option 1 is

$P_1=P_{se}+P_{ue}^2=0.015+(0.05)^2=0.015+0.0025=0.0175$

b) The probability that both suppliers will be disrupted using option 2:

$P_2=P_{se}+P_{ue}^2=0.002+(0.13)^2=0.002+0.0169=0.0189$

Pue = probability of a unique event

Pse = probability of a superevent

3 0

3 years ago

(4x2 - 10x -24) (2x + 3)

jeyben [28]

Answer:

Apply BODMAS

Step-by-step explanation:

PLEASE FIND THE PICTURE BELLOW

SOLUTION STEPS

(4⋅2−10x−24)(2x+3)

Multiply 4 and 2 to get 8.

(8−10x−24)(2x+3)

Subtract 24 from 8 to get −16.

(−16−10x)(2x+3)

Apply the distributive property by multiplying each term of −16−10x by each term of 2x+3.

−32x−48−20x

−30x

Combine −32x and −30x to get −62x.

−62x−48−20x

3 0

2 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

3 years ago