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lianna [129]
3 years ago
10

A kangaroo hops 60m to the east in 5 seconds what is the kangaroos average velocity

Mathematics
1 answer:
Archy [21]3 years ago
8 0
Average velocity is defined as displacement/time. In this case the kangaroo moves (or is displaced) 60m east and this occurs in 5 seconds, thus:
Average velocity = 60m east/5s
= 12m/s east
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Calculate, to four decimal places, the first ten terms of the sequence. an = 1 +(−4/9)^n
Lemur [1.5K]

The first ten terms of the sequence a_n=1 + (-\frac{4}{9} )^n is<em> 0.5556, 1.1975, 0.9122, 1.039, 0.9827, 1.0077, 0.9966, 1.0015, 0.9993 and 1.0003</em>

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Given that:

a_n=1 + (-\frac{4}{9} )^n\\\\When\ n=1:a_1=1 + (-\frac{4}{9} )^1=0.5556\\\\When\ n=2:a_2=1 + (-\frac{4}{9} )^2=1.1975\\\\When\ n=3:a_3=1 + (-\frac{4}{9} )^3=0.9122\\\\When\ n=4:a_4=1 + (-\frac{4}{9} )^4=1.039\\\\When\ n=5:a_5=1 + (-\frac{4}{9} )^5=0.9827\\\\When\ n=6:a_6=1 + (-\frac{4}{9} )^6=1.0077\\\\When\ n=7:a_7=1 + (-\frac{4}{9} )^7=0.9966\\\\When\ n=8:a_8=1 + (-\frac{4}{9} )^8=1.0015\\\\When\ n=9:a_9=1 + (-\frac{4}{9} )^9=0.9993\\\\

When\ n=10:a_{10}=1 + (-\frac{4}{9} )^{10}=1.0003\\

The first ten terms of the sequence a_n=1 + (-\frac{4}{9} )^n is<em> 0.5556, 1.1975, 0.9122, 1.039, 0.9827, 1.0077, 0.9966, 1.0015, 0.9993 and 1.0003</em>

Find out more on equation at: brainly.com/question/2972832

#SPJ1

3 0
2 years ago
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lys-0071 [83]

How to solve your problem

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9x−4+3x9x-4+3x9x−4+3x

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1

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