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aliya0001 [1]
3 years ago
15

The song writing competition, the minimum length of a song is 2.5 minutes. The maximum length of a song is 5.5 minutes. what is

the absolute value equation that has these minimum and maximum lengths as it's solution.
Mathematics
1 answer:
grin007 [14]3 years ago
5 0
X represents the length of the song.
x is an element of the interval 2.5, 5.5
the midpoint of the interval 2.5, 5.5 is (1/2)(2.5 + 5.5) = 8/4=4
meaning: for every x these inequalities hold:
2.5 <=x <= 5.5 subtract 4 feom each side

2.5 -4<=x -4 < = 5.5 - 4
-1.5< = x - 4 < = 1.5
or
|x - 4| < = 1.5
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Gekata [30.6K]

Answer:

We choose C

Step-by-step explanation:

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Basically, interquartile range represents the width or "dispersion" of the set. [1] The interquartile range is determined by the difference between the top quartile (25% highest) and lower quartile (25% lowest) point of the data set.

From the picture, we can find that:

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So the interquartile range of the distances for the bus drivers is 5 miles less than the interquartile range of the distances for the teachers.

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3 years ago
(6) (Bonus) Determine the Cartesian equation of the surface with spherical coordinate equation rho = 2 cos θ sin φ − 2 sin θ sin
KonstantinChe [14]

Answer:

Hence, the sphere has a radius of \sqrt{3}  and is centered at the point (1,-1,1)

Step-by-step explanation:

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We have to take into account the relation between coordinates

\rho=\sqrt{x^2+y^2+z^2}\\x=\rho cos\theta sin\phi\\y=\rho sin\theta sin\phi\\z= \rho cos\phi

by substituting we have:

\rho=2[\frac{x}{\rho}-\frac{y}{\rho}+\frac{z}{\rho}]\\\\\rho^2=2x-2y+2z\\\\x^2+y^2+z^2=2x-2y+2z

We have to complete squares:

(x^2-2x+1)+(y^2+2y+1)+(z^2-2z+1)-1-1-1=0\\\\(x-1)^2+(y+1)^2+(z-1)^2=3

Hence, the sphere has a radius of \sqrt{3}  and is centered at the point (1,-1,1)

hope this helps!!

4 0
3 years ago
What is the cube of 6
alexandr402 [8]

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Brainliest to help me reach expert? ;)

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8 0
3 years ago
Read 2 more answers
[Sin A + Sin 3A + Sin 5A + Sin 7A]÷<br><br>[Cos A + Cos 3A + Cos 5A + Cos 7A]<br>=Tan 4A<br>​
Ainat [17]

Do the following rewrites:

sin(<em>x</em>) = sin(4<em>x</em> - 3<em>x</em>) = sin(4<em>x</em>) cos(3<em>x</em>) - cos(4<em>x</em>) sin(3<em>x</em>)

sin(7<em>x</em>) = sin(4<em>x</em> + 3<em>x</em>) = sin(4<em>x</em>) cos(3<em>x</em>) + cos(4<em>x</em>) sin(3<em>x</em>)

sin(3<em>x</em>) = sin(4<em>x</em> - <em>x</em>) = sin(4<em>x</em>) cos(<em>x</em>) - cos(4<em>x</em>) sin(<em>x</em>)

sin(5<em>x</em>) = sin(4<em>x</em> + <em>x</em>) = sin(4<em>x</em>) cos(<em>x</em>) + cos(4<em>x</em>) sin(<em>x</em>)

cos(<em>x</em>) = cos(4<em>x</em> - 3<em>x</em>) = cos(4<em>x</em>) cos(3<em>x</em>) + sin(4<em>x</em>) sin(3<em>x</em>)

cos(7<em>x</em>) = cos(4<em>x</em> + 3<em>x</em>) = cos(4<em>x</em>) cos(3<em>x</em>) - sin(4<em>x</em>) sin(3<em>x</em>)

cos(3<em>x</em>) = cos(4<em>x</em> - <em>x</em>) = cos(4<em>x</em>) cos(<em>x</em>) + sin(4<em>x</em>) sin(<em>x</em>)

cos(5<em>x</em>) = cos(4<em>x</em> + <em>x</em>) = cos(4<em>x</em>) cos(<em>x</em>) - sin(4<em>x</em>) sin(<em>x</em>)

So in the numerator, we have

sin(<em>x</em>) + sin(3<em>x</em>) + sin(5<em>x</em>) + sin(7<em>x</em>)

= 2 sin(4<em>x</em>) cos(3<em>x</em>) + 2 sin(4<em>x</em>) cos(<em>x</em>)

= 2 sin(4<em>x</em>) (cos(3<em>x</em>) + cos(<em>x</em>))

In the denominator,

cos(<em>x</em>) + cos(3<em>x</em>) + cos(5<em>x</em>) + cos(7<em>x</em>)

= 2 cos(4<em>x</em>) cos(3<em>x</em>) + 2 cos(4<em>x</em>) cos(<em>x</em>)

= 2 cos(4<em>x</em>) (cos(3<em>x</em>) + cos(<em>x</em>))

So we have

(sin(<em>x</em>) + sin(3<em>x</em>) + sin(5<em>x</em>) + sin(7<em>x</em>)) / (cos(<em>x</em>) + cos(3<em>x</em>) + cos(5<em>x</em>) + cos(7<em>x</em>))

= (2 sin(4<em>x</em>) (cos(3<em>x</em>) + cos(<em>x</em>))) / (2 cos(4<em>x</em>) (cos(3<em>x</em>) + cos(<em>x</em>)))

= sin(4<em>x</em>) / cos(4<em>x</em>)

= tan(4<em>x</em>)

QED

6 0
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