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2 years ago

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What is the domain for each function below? Choose the correct inequality symbol from the drop down menu and fill in the blank w

ith the correct integer so that the appropriate domain for each is described. domain: x domain: x

1 answer:

erik [133]2 years ago

4 0

Answer:

$x \le 3$

$x \ge 0$

Step-by-step explanation:

The functions, missing from the question are:

$f(x) = \sqrt{3 - x$

$f(x) = \sqrt{3 x$

Required

Determine the domain

To get the domain, means we want to get the possible values of x.

(a)

$f(x) = \sqrt{3 - x$

Equate 3 - x to 0

$3 - x = 0$

Solve for x

$-x = -3$

$x = 3$

From the calculated value of x above;

When x exceeds 3; f(x) = undefined

So: the domain is:

$x \le 3$

(b)

$f(x) = \sqrt{3x$

Equate 3x to 0

$3x = 0$

Solve for x

$x = 0$

From the calculated value of x above;

When x falls below 0; f(x) = undefined

So: the domain is:

$x \ge 0$

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farmer ryan has corn field that is nearly a rectangle. the length is five meters more than the width. the perimeter is about 210

Tcecarenko [31]

Farmer ryan has a corn feild that is in the shape of a rectangle. The length is five meters more than the width. The perimeter is 210 meters.

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Question 1123428: A bucket is in the shape of a truncated cone.The height of the bucket is 45 cm,and the diameters of the top and the bottom surfaces are 56 cm and 14 cm respectively.Find the the quantity (in litres) of the water the bucket can hold.

Answer by Boreal(9049) About Me (Show Source):

You can put this solution on YOUR website!

V=(1/3)(r1^2+r1r2+r2^2)h*pi

=(1/3)*45*(28^2+196+49)*pi

=15(1029)*pi

A piece of paper is in the shape of a sector of a circle whose radius is 12 cm and the central angle of the sector is 120 degree.

It is rolled to form a cone of the biggest possible capacity.

Find the capacity of cone.

:

Find the total area of the sector

A = 120%2F360*pi%2A12%5E2

A = 150.8 sq/cm, this is surface area of the cone

:

Surface area (SA) of a cone without the end formula

The slant length (s) = the radius of the paper; 12 cm

pi%2Ar%2As+=+SA

pi%2Ar%2A12+=+150.8 , find r, the radius of the cone

r = 150.8%2F%2812%2Api%29

r = 4 cm is the radius

find the height of the cone

h = sqrt%28s%5E2-r%5E2%29

h = sqrt%2812%5E2-4%5E2%29

h = 11.3 cm is the height of the cone

Find the capacity (volume)

V+=+1%2F3*+pi%2Ar%5E2%2Ah

V = 1%2F3*pi%2A4%5E2%2A11.3

V = 189.33 cu/cm

6 0

3 years ago

PLS HELP I MUST DO THIS BEFORE 1:30 PLS HELP OR I WILL FAIL TY

sweet [91]

The MAD for City 1 is 24.
The MAD for City 2 is 11 1/3.

The MAD for City 2 is less than the MAD for City 1, which means the average monthly temperatures of City 2 vary less than the average monthly temperatures for City 1.

7 0

3 years ago

The measure of an angle is 8 degrees less than three times the measure of another angle. If the two angles are supplementary (tw

Kitty [74]

Answer:

The measure of the larger is $133^{\circ}$ .

Step-by-step explanation:

Given:

The measure of an angle is $8$ degrees less than three times the measure of another angle.

The two angles are supplementary.

To find: The measure of the larger angle.

Solution:

Let the measure of the smaller angle be $x^{\circ}$ .

Then the measure of the larger angle be $3x^{\circ}-8^{\circ}$ .

The two angles are supplementary, so their sum is $180^{\circ}$ .

So, $x^{\circ}+3x^{\circ}-8^{\circ}=180^{\circ}$

$\Rightarrow 4x^{\circ}-8^{\circ}=180^{\circ}$

$\Rightarrow 4x^{\circ}=180^{\circ}+8^{\circ}$

$\Rightarrow 4x^{\circ}=188^{\circ}$

$\Rightarrow x^{\circ}=\frac{188^{\circ}}{4}$

$\Rightarrow x^{\circ}=47^{\circ}$

So, the measure of the smaller angle is $47^{\circ}$ .

And, the measure of the larger angle is $3\times47^{\circ}-8^{\circ}=133^{\circ}$ .

Hence, the measure of the larger is $133^{\circ}$ .

3 0

3 years ago

The length of a rectangular dog park is 50 feet. The width is half the length. Decide if each set of steps could be used to find

Leokris [45]

Answer:

4. Find the Width by dividing 50 by 2. Then double the length, double the width, and add the products to find the perimeter = Yes

Step-by-step explanation:

The length of a rectangular dog park is 50 feet. The width is half the length.

The formula for Perimeter is given as:

P = 2L + 2W

Length = 50 feet

Width = 50 feet/2

= 25 feet

P = 2(50) + 2(25)

P = 100 + 50

P = 150 feet

Hence, writing yes or no in front of each step in the options.

1. Find the Width by dividing 50 by 2. Then add the length and width and multiply the sum to find the perimeter = No

2. Find the Width by dividing 50 by 2. Then multiply the length by the width to find the perimeter. = No

3. Find the Width by dividing 50 by 2. Then add the length and the width to find the perimeter.= No

4. Find the Width by dividing 50 by 2. Then double the length, double the width, and add the products to find the perimeter = Yes

5. Find the Width by dividing 50 by 2. Then add the sum of each length and the sum of each width to find the perimeter = No

THE CORRECT STEP IN THE ABOVE OPTION IS OPTION 4

4 0

2 years ago

Consider the differential equation

Ainat [17]

Answer:

$W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )=e^x$

Step-by-step explanation:

Let $y=e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y'=\frac{1}{2}e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y''=\frac{1}{4}e^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (\frac{1}{4}e^{\frac{x}{2}} \right )-4\left ( \frac{1}{2}e^{\frac{x}{2}}\right )+e^{\frac{x}{2}}\\=e^{\frac{x}{2}}-2e^{\frac{x}{2}}+e^{\frac{x}{2}}\\=2e^{\frac{x}{2}}-2e^{\frac{x}{2}}\\=0$

So, $y=e^{\frac{x}{2}}$ is the solution of the given equation.

Now, let $y=xe^{\frac{x}{2}}$

Differentiate with respect to $x$

$y'=e^{\frac{x}{2}}+\frac{x}{2}e^{\frac{x}{2}}=e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )$

Differentiate with respect to $x$

$y''=\frac{1}{2}e^{\frac{x}{2}}+\frac{1}{2}e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )=e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}} \right )-4\left ( e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )\right )+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-2e^{\frac{x}{2}}(2+x)+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-4e^{\frac{x}{2}}-2xe^{\frac{x}{2}}+xe^{\frac{x}{2}}\\=0$

To find: $W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )$

Solution:

Let $u=e^{\frac{x}{2}}\,,\,v=xe^{\frac{x}{2}}$

$W(u,v)=\left | \begin{matrix}u&v\\u'&v' \end{matrix} \right |\\=\left | \begin{matrix}e^{\frac{x}{2}}&xe^{\frac{x}{2}}\\\frac{1}{2}e^{\frac{x}{2}}&e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \end{matrix} \right |\\=e^{\frac{x}{2}}\left [ e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \right ]-\frac{1}{2}e^{\frac{x}{2}}xe^{\frac{x}{2}}\\=e^x\left ( 1+\frac{x}{2} \right )-\frac{1}{2}xe^x\\=e^x+\frac{1}{2}xe^x-\frac{1}{2}xe^x\\=e^x$

5 0

2 years ago