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Veseljchak [2.6K]

3 years ago

12

Daniel had $25 to spend at the fair. If the admission to the fair is $4 and the rides cost $1.50 each, what is the greatest numb

er of rides Daniel can go on?

2 answers:

mote1985 [20]3 years ago

8 0

25= 1.50x+4
Subtract 4
21= 1.50x
Divide by 1.50
X=14 rides or less

Norma-Jean [14]3 years ago

7 0

The greatest number of rides he can go = ($25-$4)/$1,50=14

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The Florida Fish and Wildlife Conversation Commission (FWC) plans to hire hunters to kill the Burmese python, an invasive specie

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Complete Question

The complete question is shown on the first uploaded image

Answer:

FWOC should hire 70,000 hunters to completely remove the python in the area

Step-by-step explanation:

From the question we are given that the Initial Pythons $P_o=$ 100,000

And the final python $P_f$ $=100,000+\frac{P}{10,000}(100,000-P)$

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For Max Number of Python

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Using Mathematical relations

from the Question

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Read 2 more answers

Area of a Triangle

jek_recluse [69]

Answer:

3.51 (round to the nearest hundreds)

Step-by-step explanation:

PART A

slope of the line passing through AC : (6-1)/(1-2) = 5/-1 = -5

equation of the line passing through B and perpendicular to AC:

y-3 = 1/5(x-3)

y-3 = 1/5x -3/5

y = 1/5 x - 3/5 + 3

y = 1/5 x + (-3+15)/5

y = 1/5 x + 12/5

PART B

line that passes through the points A and C

y-1 = -5(x-2)

y = -5x+10 + 1

y = -5x + 11

Interception of the lines

y = 1/5x + 12/5

y = -5x + 11

-5x + 11 = 1/5x + 12/5

-25x +55 = x + 12

26x = 43

x = 43/26

y = -5(43/26) + 11

y= -215/26 + 11

y = (-215+286)/26 = 71/26

D (1.65, 2.73)

PART C

AC = $\sqrt{(1-2)^2 + (6-1)^2} = \sqrt{1 + 25} = \sqrt{26}$ = 5,09902

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PART D

AREA = (5,09902 * 1.376735)/2 = 3,509999

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3 years ago

wind resistance varies jointly as an objects surface area and velocity. if an object traveling at 55 miles per hour with a surfa

nikdorinn [45]

Answer:

25 miles per hour

Step-by-step explanation:

Given

$W = Wind\ resistance$

$A = Surface\ Area$

$V = Velocity$

The joint variation can be represented as:

$W\ \alpha\ A*V$

Where:

$V = 55; A = 20; W = 220$

Required

Find V, when: $A = 55; W = 275$

We have:

$W\ \alpha\ A*V$

Express as an equation

$W= k *A*V$

Where k is the constant of variation

Make k the subject

$k = \frac{W}{A*V}$

When: $V = 55; A = 20; W = 220$

We have:

$k = \frac{220}{20 *55 }$

$k = \frac{220}{1100}$

$k = 0.2$

When: $A = 55; W = 275$

We have:

$k = \frac{W}{A*V}$

Substitute: $A = 55; W = 275$ and $k = 0.2$

$0.2 = \frac{275}{55 * V}$

Make V the subject

$V= \frac{275}{55 * 0.2}$

$V= \frac{275}{11}$

$V= 25$

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3 years ago