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Bumek [7]
3 years ago
13

Suppose you walk down a school hallway at a rate of 5 ft/s, and you reach your destination in 20 seconds. How do you write a fun

ction that models this situation, and how do you determine the domain?
Mathematics
2 answers:
8090 [49]3 years ago
7 0

Answer:

Step-by-step explanation:

Let's write a equation.

5 ft = 20 seconds

(five feet is reachable in 20 seconds)

Divide both sides by 5.

5/5=1

20/5=4

1 ft = 4 seconds

Let's say the destination is d and f is foot.

d = 4f

KATRIN_1 [288]3 years ago
7 0
Why would I walk down a school hallway
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Abcdef is a regular hexagon what is its perimeter what is its area
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Answer:

p=36

a=3*\sqrt{3}*18

Step-by-step explanation:

p=sum of all sides

a=(3\sqrt{3}*s^{2})/2

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The length of a rectangle is 4 more than twice the width. If the length is 62, find the width.
lukranit [14]
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3 0
3 years ago
A ladder is placed 30 inches from a wall. It touches the wall at a height of 50 inches from the ground.
tankabanditka [31]
To find the length of the ladder, you need to do Pythagorean theorem.
50^2 + 30^2 = x^2
2500 + 900 = x^2
x^2 =3400
x= square root of 3400
58.31 OR 10 root34

To find the angle:
tan theta = opposite/adjacent
50/30 = 5/3
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= 59.04


8 0
3 years ago
Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths. X2 y2 − 2x 2y −
Colt1911 [192]

Circles (represented by their equations in general form) in ascending order of their radius lengths can be arranged as first-fifth-second-forth-third-seventh-sixth.

<h3>What is the equation of circle?</h3>

The equation of the circle is the equation which is used to represent the circle in the algebraic equation form with the value of center point in the coordinate plane and measure of radius.

The standard form of the equation of the circle can be given as,

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

Here (h,k) is the center of the circle and (r) is the radius of the circle.

Lets arrange the first equation, in standard form of equation of circle as,

x^2 +y^2 - 2x+ 2y -1 = 0 \\(x^2 - 2x) +(y^2+ 2y )= 1\\(x^2 - 2x+1) +(y^2+ 2y +1)= 1+1+1\\(x-1)^2+(y+1)^2=(\sqrt{3})^2

The radius of this circle is √(3).

Lets arrange the second equation, in standard form of equation of circle as,

x^2 +y^2 - 4x+ 4y - 10 = 0  \\(x^2 - 4x) +(y^2+ 4y )= 10\\(x^2 - 4x+4) +(y^2+ 4y +4)= 10+4+4\\(x-2)^2+(y+2)^2=(\sqrt{18})^2

The radius of this circle is √(18).

Lets arrange the third equation, in standard form of equation of circle as,

x^2 +y^2 - 8x - 6y - 20 = 0   \\(x^2 - 8x) +(y^2-6y )= 20\\(x^2 - 8x+16) +(y^2- 6y +9)= 10+16+9\\(x-4)^2+(y-3)^2=(\sqrt{45})^2

The radius of this circle is √(45).

Similarly, radius of the forth, fifth, sixth and seventh circle is √(23), √(5), √(117) and √(46) units.

Hence, the circles (represented by their equations in general form) in ascending order of their radius lengths can be arranged as first-fifth-second-forth-third-seventh-sixth.

Learn more about the equation of circle here;

brainly.com/question/1506955

3 0
2 years ago
Read 2 more answers
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