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3 years ago

15

A data set contains an independent and a dependent variable. Which must be true of the data set if a linear function can be used

to represent the data?

1 answer:

Andrews [41]3 years ago

4 0

Answer:

the set must have a constant additive rate of change. the set must have a multiplicative rate of change.

Step-by-step explanation:

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Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

<u>apply chain rule</u>

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$ <u />

<u>maximum volume:</u> <u>when h = 40/3</u>

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

<u>minimum volume:</u> <u>when h = 0</u>

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

6 0

2 years ago

Read 2 more answers

If f(x)=3x-1 and g(x)=x+2, find (f-g)(x)

siniylev [52]

Answer:

(f - g)(x) = 2x - 3

Step-by-step explanation:

(f - g)(x)

= f(x) - g(x)

= 3x - 1 - (x + 2)

= 3x - 1 - x - 2

= 2x - 3

7 0

2 years ago

Manuel and Joe order pasta for $4.55, salad for $5.15, and 2 glasses of lemonade for $1.70 each. The tax is $1.05. How much chan

Svetradugi [14.3K]

They should get 85 cents. ($0.85)

7 0

3 years ago

A line passes through the points (-6, 4) and (-2, 2). Which is the equation of the line?

cupoosta [38]

Answer:

$\displaystyle y=-\frac{1}{2}x+1$

Step-by-step explanation:

The equation of any line in slope-intercept form is:

y=mx+b

Being m the slope and b the y-intercept.

Assume we know the line passes through points A(x1,y1) and B(x2,y2). The slope can be calculated with the equation:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Two points are given: (-6,4) and (-2,2). Calculating the slope:

$\displaystyle m=\frac{2-4}{-2+6}=\frac{-2}{4}=-\frac{1}{2}$

The equation of the line is, so far:

$\displaystyle y=-\frac{1}{2}x+b$

To calculate the value of b, we use any of the given points, for example (-6,4):

$\displaystyle 4=-\frac{1}{2}(-6)+b$

$\displaystyle 4=3+b$

Solving:

b = 1

The equation of the line is:

$\boxed{\displaystyle y=-\frac{1}{2}x+1}$

We can see none of the choices is correct.

8 0

3 years ago

A) make U subject of the formula E= m/2g(v^2-u^2)

ollegr [7]

Answer:

u = v2−2Egm−−−−−−−√

Step-by-step explanation:

2Eg = 2g (M2g(V2−U2)

2Eg = m(V2 - U2)

2Eg - mV2 - mU2

mU2 = mV2 - 2Eg

mU2m=mV2−2Egm

U2 = mV2−2Egm

= mV2m−2Egm

U2 = V2 - 2Egm

U = V2−2Egm−−−−−−−−√

7 0

3 years ago