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Rainbow [258]
3 years ago
12

On a blueprint of a house, 48 millimeters represents 7 meters. The length of the living room is 60 millimeters on the blueprint.

what is the actual length
Mathematics
1 answer:
Pani-rosa [81]3 years ago
3 0
Here is the equation 48/7=60/x, 48÷7=6.8571428571. Now we know that every one meter in real life is equal to 6.8571428571 millimeters so, 60÷6.8571428571= 8.75 meters. The living room is 8.75 meters
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1. Calculate the acceleration of riding her bicycle in a straight line that speeds up from 4 m/s to 12 m/s in 10 seconds. SOLUTI
valentina_108 [34]
<h3>✒️ACCELERATION</h3>

=================================

\large\sf\underline{Problem:}

Calculate the acceleration of riding her bicycle in a straight line that speeds up from 4 m/s to 12 m/s in 10 seconds.

=================================

\large\sf\underline{Answer:}

\qquad \qquad \huge \rm = 0.8 \: m/s²

=================================

\large\sf\underline{Solution:}

Here we have a given of,

  • \tt{(12\:m/s) \:= \:V_f = (final\: velocity) }
  • \tt{(4\:m/s)\:=\:V_i = (initial \: velocity) }
  • \tt{(10\:sec) \:=\:T = (elapsed time) }

\bold{Formula\:||\: Acceleration = \frac{v\: - \:v_0}{t} = \frac{ΔV}{ΔT} }

Because acceleration has a direction <u>(</u><u>riding her bicycle in a straight line</u><u>)</u>, it is important to always subtract the final velocity represented as \bold{V_f} from your initial velocity also represented as \bold{V_i}.

Now 'Work it out' and solve for Acceleration of riding her bicycle using the given formula.

  • \large\rm{A = \frac{V_f \:- \:V_i}{T} }
  • \large\rm{A = \frac{(12\:m/s\:-\:4\:m/s) }{10\:sec}}
  • \large\rm{A = \frac{ 8\:m/s}{10\:sec} }
  • \large\rm{A = 0.8\:m/s²}

Hence, The total acceleration of the girl riding her bicycle is 0.8 meters per second squared (m/s²).

=================================

૮₍´˶• . • ⑅ ₎ა

3 0
2 years ago
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liubo4ka [24]

Answer:

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Step-by-step explanation:

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*View link below for more comprehensive information.

https://www.ncbi.nlm.nih.gov/books/NBK114332/

4 0
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Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = (x2 − 1)3, [−1, 2]
Alja [10]

Given :

F(x)=3(x^2-1)

To Find :

the absolute minimum and absolute maximum values of f on the given interval

[-1,2] .

Solution :

Now , getting first order differential equation and equating its equal to zero.

\dfrac{d(3x^2-3)}{dx}=0\\\\6x=0\\x=0

So , x=0 is critical point .

Now , coefficient of x^2 is positive .

Therefore , it is increasing function after  x=0 .

So , min value will be at , x=0.

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And maximum value will be the maximum at the x=2 because it is increasing function .

Max=(2^2-1)\times 3=9

Therefore , max and min value is 9 and -3 respectively .

Hence , this is the required solution .

3 0
3 years ago
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