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3 years ago

How can you choose values for x when making a table of value representing a real world situation?

Mathematics

1 answer:

coldgirl [10]3 years ago

7 0

You can simply count 0,1,2,3 when zero is your starting point

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A rectangle has a length of x+2, a width of 7, and a perimeter of 32. The value of x is ?

beks73 [17]

Perimeter = sum of all sides

32 = 2(x+2) + 2(7)
32 = 2x+4 + 14
32 = 2x+18
14 = 2x
/2 /2

7 = x

4 0

3 years ago

Read 2 more answers

Robin and Meg are traveling from New York to Florida, and plan to stop at three different campgrounds along the way. After choos

matrenka [14]

You are correct. For each of the three possibilities for their second day, there are two possibilities for the third day.

The total number of possibilities is 3*2 = 6.

6 0

2 years ago

The volume of the pyramid shown in the figure is cubic centimeters. If the slant height of the pyramid increases by 4 centimeter

Nostrana [21]

Answer:

Part a) The volume of the original pyramid is $15\ cm^{3}$

Part b) The volume of the pyramid increases by $6\ cm^{3}$

Step-by-step explanation:

<u><em>The complete question is</em></u>

Part 1) The volume of the pyramid shown in the figure is 9,15,21, or 63 cubic centimeters?. Part 2) If the slant height of the pyramid increases by 4 centimeters and its height increases by 2 centimeters, the volume of the pyramid increases by 6,9,12 or 21 cubic centimeters?

we know that

The volume of the pyramid is equal to

$V=\frac{1}{3}Bh$

where

B is the area of the base

h is the height of pyramid

see the attached figure to better understand the problem

Step 1

Find the volume of the original pyramid

the area of the base B is equal to

$B=3^{2}=9\ cm^{2}\\h=5\ cm$

substitute

$V=\frac{1}{3}(9)(5)=15\ cm^{3}$

Step 2

Find the volume of the new pyramid

$B=9\ cm^{2}$ -------> the area of the base is the same

$h=5+2=7\ cm$ ------> the height increase by

substitute

$V=\frac{1}{3}(9)(7)=21\ cm^{3}$

Subtract the original volume from the new volume

$21\ cm^{3}-15\ cm^{3}=6\ cm^{3}$

5 0

2 years ago

Calcular los perimetros de las siguientes figuras porfa ayúdenme ya

Basile [38]

Answer:

solo suma todos los valores x+4(3)+(x+x)

en el segundo x+5(2)+x+3(2)

en el tercero x-3+(2x+5)+(2x+5)

y en el cuarto x(3)+(x+4)

7 0

3 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