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2 years ago

What values make the inequality v<4 true.

Mathematics

1 answer:

VLD [36.1K]2 years ago

6 0

Answer:

all except "4" "4.001" "4.1" "4.01" "5" "7" "9" "12"

Step-by-step explanation:

4 is exculded because its u<4 not u<=4

have a good day !

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Write the equation which represents the following transformations

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Step-by-step explanation:

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2 years ago

Un jardín rectangular de 50 cm de largo por 34 m de ancho está rodeado por un camino de arena uniforme.Halla la anchura de dicho

Anastasy [175]

Answer:

5.85 m

Step-by-step explanation:

The width of the sand road can be calculated knowing its area and the dimensions of the rectangular garden as follows:

$A_{g} = a.b$

<u>Where:</u>

Ag: is the area of the rectangular garden

a: is the length of the rectangular garden = 50 cm = 0.5 m

b: is the width of the rectangular garden = 34 m

$A_{s} = 540 m^{2}$

<u>Where</u>:

As: is the area of the sand road

The relation between the area of the sand road and the area of the rectangular garden is the following:

$A_{s} + A_{g} = (a+2x)*(b+2x)$

$540 m^{2} + 0.5m*34m = ab + 2ax + 2bx + 4x^{2}$

$557 m^{2} = ab + 2x(a + b) + 4x^{2}$

$557 m^{2} - 17m^{2} - 2x(34.5 m) - 4x^{2} = 0$

$540 - 69x - 4x^{2} = 0$

By solving the above equation for x we have two solutions:

x₁ = -23.10 m

x₂ = 5.85 m

Taking the positive value, we have that the width of the sand road is 5.85 m.

I hope it helps you!

7 0

2 years ago

Atown uses positive numbers to track increases in population and negative numbers to track decreases in population.

amid [387]

Answer:

Step-by-step explanation:

-60

Step-by-step explanation:

Increase in population is given by positive sign

Change in population in 5 years = -300

Decrease in population is represented by negative sign

Change in population in 1 year = -300\5

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Town's population faces a decrease of 60 residents each year.

4 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

2 years ago