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3 years ago

How do I find the value of x

Mathematics

2 answers:

Mice21 [21]3 years ago

5 0

Answer:

2.6

Step-by-step explanation:

A^2 + B^2 =C^2

1^2+2.4^2=6.76

Then you find the square root of 6.76 which is 2.6

Hope it helps

IgorC [24]3 years ago

3 0

Answer:

The value of x is 2.6

Step-by-step explanation:

Use pythagorean theorem

a^2+b^2=c^2

Then substitute

1^2+2.4^2=c^2

Then evaluate

1+5.76=c^2

Add

c^2=6.76

Isolate c by finding the sqrt of 6.76

c=2.6

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

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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 derivative of the position, 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:

$6t - 6 = 0$

$6t = 6$

$t = \frac{6}{6}$

$t = 1$

Hence 1 second.

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

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2 years ago

Alan says that the distance from Hadley to Greenville is 16 miles. Is he correct?Explain

yuradex [85]

Answer:

Step-by-step explanation:

Considering the complete question attached in figure.

Consider

Hadley = x

Parker = y

Greenville = z

Distance between Greenville and Parker = yz = 29

Given that distance between Hadley and Greenville is 13 miles less than the distance between Parker and Greenville.

xz = yz -13 --- (1)

Distance between hardley and greenville = xz = ?

From (1)

xz = 29 -13

xz = 16 miles

So Alan is right that distance between Hadley and Greenville is 16 miles

8 0

3 years ago