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3 years ago

What is the average rate of change of f(x) = -x2 + 3x + 6 over the interval –3

Mathematics

1 answer:

Rufina [12.5K]3 years ago

3 0

Answer:

$\frac{\Delta y}{\Delta x} =\frac{f(x_2)-f(x_1)}{x_2-x_1} =\frac{6-(-12)}{3-(-3)} =\frac{18}{6}= 3$

Step-by-step explanation:

To find the average rate of change of a function over a given interval, basically you need to find the slope. The mathematical definition of the slope is very similar to the one we use every day. In mathematics, the slope is the relationship between the vertical and horizontal changes between two points on a surface or a line. In this sense, the slope can be found using the following expression:

$Average\hspace{3}rate\hspace{3}of\hspace{3}change=Slope=\frac{y_2-y_1}{x_2-x_1} =\frac{f(x_2)-f(x_1)}{x_2-x_1}$

So, the average rate of change of:

$f(x)=-x^2+3x+6$

Over the interval $-3$

Is:

$f(x_2)=f(3)=-(3)^2+3(3)+6=-9+9+6=6\\\\f(x_1)=f(-3)=-(-3)^2+3(-3)+6=-9-9+6=-12$

$\frac{\Delta y}{\Delta x} =\frac{f(x_2)-f(x_1)}{x_2-x_1} =\frac{6-(-12)}{3-(-3)} =\frac{18}{6}= 3$

Therefore, the average rate of change of this function over that interval is 3.

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Find the volume of the rectangular prism. Express your answer as a simplified mixed number.

Vedmedyk [2.9K]

Answer:

414 9/16 m³

$414\frac{9}{16}m^{3}$

Step-by-step explanation:

The volume of a Rectangular prism is calculated as Length × Width × Height

From the attached diagram,we can see that

The length = 8 3/8 m

The width = 6 m

The height =8 1/4 m

Converting the mixed fraction to improper fraction in other to simplify it better

The length = 67/8 m

The height = 33/4 m

The volume of the rectangular prism =

67/8 m × 6m × 33/4m

= (13266/32)m³

= 6633/16 m³

= 414 9/16 m³

= $414\frac{9}{16}m^{3}$

3 0

3 years ago

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

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2 years ago

Who needs special prayer

kupik [55]

Answer:

me, please me

Step-by-step explanation:

I'm going through court cases because my dad raped me and so on

8 0

3 years ago

Read 2 more answers

What isn't the bets approximation for the area of this circle labeled 8 inches use3.14 to approximate pi

Arte-miy333 [17]

Answer:

50.272 in^2

Step-by-step explanation:

Step one:

Given data

Circles are described using the diameter parameter

Diameter of the circle is 8 inches

radius = 4 inche= d/2

Step two:

The area of the circle is

A= πr^2

A= 3.142*4^2

A=3.142*16

A=50.272 in^2

8 0

3 years ago

5 (RootIndex 3 StartRoot x EndRoot) + 9 (RootIndex 3 StartRoot x EndRoot)

Ratling [72]

Answer:

C. 14 RootIndex 3 StartRoot x EndRoot)

Step-by-step explanation:

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8 0

2 years ago