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KiRa [710]
3 years ago
8

There were 567 people at a concert when a band started playing. After each song, only one-third of the people stayed to hear the

next song. How many people were left at the concert after x songs? Write a function to represent this scenario.
Mathematics
2 answers:
emmasim [6.3K]3 years ago
8 0
To write the function correctly, it is important to assign variables correctly and understand the situation of the problem clearly. For this, we let y the number of people and x as the number of songs played.

At x = 0   y = 567
at x = 1   y = 567 - 567(1/3)
at x = 2   y = 567 - 567(1/3)(1/3)
at x = 3   y = 567 - 567(1/3)(1/3)(1/3)

Therefore, the number of people left after x songs would be represented by the equation:

y = 567 - 567(1/3)x
y = 567 ( 1- x/3 )
vladimir2022 [97]3 years ago
8 0

Answer:

For Apex the answer is f(x)=567(1/3)^x

Step-by-step explanation:


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Because 400 is a lot and if it's 100% is 400 right so we do 80% what would that be? we don't know right.

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Please explain.
DENIUS [597]
If you want to solve this i suggest rather dividing or multiplying.
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Complete the following sentence. The slope of the graph of y = 5 x is ___
Y_Kistochka [10]

Answer:

We conclude that the slope of the graph of y = 5 x is 5.

Step-by-step explanation:

We know the slope-intercept form of the line equation

y = mx+b

where

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Given the line

y = 5x

comparing with the slope-intercept form of the line equation

The slope of the line = m = 5

Therefore, we conclude that the slope of the graph of y = 5 x is 5.

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Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes
Darina [25.2K]

Answer:

Orthocentre (intersection of altitudes) is at (37/10, 19/5)

Step-by-step explanation:

Given three vertices of a triangle

A(-5, -2)

B(7, -5)

C(3, 1)

Solution A by geometry

Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4

Slope of line normal to AB, nab = -1/(-1/4) = 4

Altitude of AB = line through C normal to AB

(y-yc) = nab(x-xc)

y-1 = (4)(x-3)

y = 4x-11           .........................(1)

Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2

Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3

Altitude of BC

(y-ya) = nbc(x-xa)

y-(-2) = (2/3)(x-(-5)

y = 2x/3 + 10/3 - 2

y = (2/3)(x+2)    ........................(2)

Orthocentre is at the intersection of (1) & (2)

Equate right-hand sides

4x-11 = (2/3)(x+2)

Cross multiply and simplify

12x-33 = 2x+4

10x = 37

x = 37/10  ...................(3)

substitute (3) in (2)

y = (2/3)(37/10+2)

y=(2/3)(57/10)

y = 19/5  ......................(4)

Therefore the orthocentre is at (37/10, 19/5)

Alternative Solution B using vectors

Let the position vectors of the vertices represented by

a = <-5, -2>

b = <7, -5>

c = <3, 1>

and the position vector of the orthocentre, to be found

d = <x,y>

the line perpendicular to BC through A

(a-d).(b-c) = 0                          "." is the dot product

expanding

<-5-x,-2-y>.<4,-6> = 0

simplifying

6y-4x-8 = 0 ...................(5)

Similarly, line perpendicular to CA through B

<b-d>.<c-a> = 0

<7-x,-5-y>.<8,3> = 0

Expand and simplify

-3y-8x+41 = 0 ..............(6)

Solve for x, (5) + 2(6)

-20x + 74 = 0

x = 37/10  .............(7)

Substitute (7) in (6)

-3y - 8(37/10) + 41 =0

3y = 114/10

y = 19/5  .............(8)

So orthocentre is at (37/10, 19/5)  as in part A.

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