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Vesnalui [34]
3 years ago
15

Laura has a step of 50cm She walks along by taking two steps forward and one step back. What is the least number of steps counti

ng forward and backwards step, she takes to reach a step 20m away? ~Thanks - don't know how to solve :(

Mathematics
2 answers:
emmasim [6.3K]3 years ago
4 0
I wrote the answer somewhere else and it wouldn’t let me copy and paste so here’s the screenshot

Vilka [71]3 years ago
3 0

Answer:

The least number of steps she takes to reach 20 m is 118 steps

Step-by-step explanation:

For Laura, the length of a walking step = 50 cm = 0.5 m

The repeating number of steps she takes = Two steps forward one step backwards

The number of forward steps it would take for her to reach 20 m = (20 m)/(0.5 m) = 40 steps

Given that she always takes two steps forward and one step backwards, we have;

The number of steps forward every three steps = 2 step forward + (-1) step (backward)

The number of steps forward every three steps = 2 - 1 = 1 step forward

To reach 39 steps forward, she would need 39 × 3 =  117 steps

To get to the 40 steps she needs just a step forward, making the total number of steps to reach 40 steps = 117 + 1 = 118 steps

Therefore, the least number of steps she takes to reach 20 m = 118 steps.

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Answer:

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Step-by-step explanation:

Partitioning between two points A and B in the ratio a:b.

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3 years ago
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Answer:

\large\boxed{-\bigg[(x-3)^2+2x\bigg]+1}\\\\\boxed{-x^2+4x-8}

Step-by-step explanation:

(g\ \circ\ f)(x)=g\bigg(f(x)\bigg)

f(x)=(x-3)^2+2x\\\\g(x)=-x+1\\\\g\ \circ\ f\to\text{put f(x) instead of x in the function g(x)}:\\\\(g\ \circ\ f)(x)=-\bigg[\underbrace{(x-3)^2+2x}_{x}\bigg]+1

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-----------------------------------

-x^2+4x-8=-x^2+4x-2^2+2^2-8=-(x^2-4x+2^2)+4-8\\\\=-(\underbrace{x^2-(2)(x)(2)+2^2}_{(a-b)^2=a^2-2ab+b^2})-4=-(x-2)^2-4=-(x+1-3)^2-4

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