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yulyashka [42]
2 years ago
11

0-6

Mathematics
1 answer:
dybincka [34]2 years ago
7 0
<h3>Given :-</h3>
  • Pranav ran 20.3 km more than Branda

  • Pravin ran 38.6 km

\\  \\

To find :

  • No. of kilometers did Brenda run.

\\  \\

<h3>Let :</h3>

  • No. of kilometers ran by Brenda be x.

\\  \\

<h3>Solution:</h3>

\\  \\

Equation formed:-

\\  \\

Total distance covered by Pravan = More distance covered by Pravan + distance covered by Brenda.

Therefore:-

\\  \\

\leadsto \sf38.6 = 20.3 + x

Write the equation

\\  \\

\leadsto \sf38.6 - 20.3 =x

When we transfer 20.3 to left side the positive sign (+) will change into negative sign (–)

\\  \\

\leadsto \sf x = 38.6 - 20.3

Arrange the equation because x is always represented at left side.

\\  \\

\leadsto \boxed {\pmb{\sf x = 18.3}}\star

After subtracting 38.6 with 20.3 we will get result as 18.3 .

\\  \\

\therefore \red{ \underline{  \pmb{\frak{Distance  ~ covered ~by ~Brenda~ is ~equal ~to~18.3~kilometers}}}}

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See below

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| -5| = 5  

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What is the value of the expression?<br><br> [(1+5)⋅2−5]⋅2
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5 0
3 years ago
A body moves s metres in a time t seconds so that s = t3 – 3t2 + 8. Find:
Lady bird [3.3K]

Using derivatives, it is found that:

i) v(t) = 3t^2 - 6t

ii) 9 m/s.

iii) a(t) = 6t - 6

iv) 6 m/s².

v) 1 second.

<h3>What is the role of derivatives in the relation between acceleration, velocity and position?</h3>
  • The velocity is the derivative of the position.
  • The acceleration is the derivative of the velocity.

In this problem, the position is:

s(t) = t^3 - 3t^2 + 8

item i:

Velocity is the <u>derivative of the position</u>, hence:

v(t) = 3t^2 - 6t

Item ii:

v(3) = 3(3)^2 - 6(3) = 27 - 18 = 9

The speed is of 9 m/s.

Item iii:

Derivative of the velocity, hence:

a(t) = 6t - 6

Item iv:

a(2) = 6(2) - 6 = 6

The acceleration is of 6 m/s².

Item v:

t for which a(t) = 0, hence:

6t - 6 = 0

6t = 6

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t = 1

Hence 1 second.

You can learn more about derivatives at brainly.com/question/14800626

7 0
2 years ago
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