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Sergio [31]
3 years ago
7

Suppose that a butterfly can fly 82 feet in 4 seconds. A dragonfly can fly 50 feet in 2 seconds. Which can fly faster and by how

much?
A) The butterfly is 4.5 feet per second faster.

B) The dragonfly is 4.5 feet per second faster.

C) dragonfly is 20.5 feet per second faster.

D) The butterfly is 24 feet per second faster.
Mathematics
2 answers:
Vilka [71]3 years ago
7 0
I belive B. Is your answer
Lady bird [3.3K]3 years ago
5 0

A butterfly can fly 82 feet in 4 seconds.

Distance traveled by butterfly = 82 feet.

Time taken by butterfly =4 seconds

Speed = distance /time

Speed of butterfly = 82/4 = 20.5 feet/second

A dragonfly can fly 50 feet in 2 seconds.

Distance traveled by dragonfly =50 feet.

Time taken by dragonfly =2 seconds.

Speed of dragonfly = 50/2 = 25 feet/second

So dragonfly can fly faster than the butterfly.

Difference in speed = speed of dragonfly - speed of butterfly = 25-20.5 = 4.5

So The dragonfly is 4.5 feet per second faster than butterfly.

Answer: B) The dragonfly is 4.5 feet per second faster.  


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mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi
chubhunter [2.5K]

Hello!  

We have the following data:  

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)  

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

a_n = a_1 + (n-1)*r

* second year salary

a_2 = a_1 + (2-1)*300

a_2 = 32000 + 1*300

a_2 = 32000 + 300

\boxed{a_2 = 32300}

* third year salary

a_3 = a_1 + (3-1)*300

a_3 = 32000 + 2*300

a_3 = 32000 + 600

\boxed{a_3 = 32600}

* fourth year salary

a_4 = a_1 + (4-1)*300

a_4 = 32000 + 3*300

a_4 = 32000 + 900

\boxed{a_4 = 32900}

* fifth year salary

a_5 = a_1 + (5-1)*300

a_5 = 32000 + 4*300

a_5 = 32000 + 1200

\boxed{a_5 = 33200}

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

a_6 = a_1 + (6-1)*300

a_6 = 32000 + 5*300

a_6 = 32000 + 1500

\boxed{a_6 = 33500}

* seventh year salary

a_7 = a_1 + (7-1)*300

a_7 = 32000 + 6*300

a_7 = 32000 + 1800

\boxed{a_7 = 33800}

*  eighth year salary

a_8 = a_1 + (8-1)*300

a_8 = 32000 + 7*300

a_8 = 32000 + 2100

\boxed{a_8 = 34100}

* ninth year salary

a_9 = a_1 + (9-1)*300

a_9 = 32000 + 8*300

a_9 = 32000 + 2400

\boxed{a_9 = 34400}

*  tenth year salary

a_{10} = a_1 + (10-1)*300

a_{10} = 32000 + 9*300

a_{10} = 32000 + 2700

\boxed{a_{10} = 34700}

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

*  eleventh year salary

a_{11} = a_1 + (11-1)*300

a_{11} = 32000 + 10*300

a_{11} = 32000 + 3000

\boxed{\boxed{a_{11} = 35000}}\end{array}}\qquad\checkmark

Respuesta:

In the eleventh year of salary he will earn more than 34700, in the case, this value will be 35000

________________________

¡Espero haberte ayudado, saludos... DexteR! =)

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3 years ago
The family car runs out of gas. You have to push it up the driveway into a service station. The ramp is 4 meters long and rises
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Answer:

Step-by-step explanation:

slope, m = 0.1m/4m = 0.025

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actual effort = 300 N

efficiency

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= 225 N / 300 N = 0.75

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Answer:

~~~~~~~~~~~Hello!~~~~~~~~~~~

Your answer should be:

<em>The number of miles Andrew's father drives on a tank of gas ≤ 350.</em>

Step-by-step explanation:

I hope this helped!

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<h2>Answer:</h2>

\boxed{15.4z-12}

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We combine like terms in order to simplify expression. To do so, we combine terms with the same variable and exponents. In this exercise, let's apply distributive property first:

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3 years ago
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Answer:   On the eighth day of diet both animals will be having same  calories.

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