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Temka [501]
3 years ago
12

A dolphin is swimming 4 feet below the surface of the water. the number -4 represents the dolphins position relative to the surf

ace of the water. what position does the number 4 represent?
A. 0 feet below water
B. 4 feet above water
C. 8 feet below the water
D. 8 feet above the water
Mathematics
1 answer:
IRISSAK [1]3 years ago
6 0
Answer: option B. 4 feet above water.

Explanation:

This is an example of the use of the integer numbers.

Integer numbers are positive, zero and negative numbers.

In this case, zero is used to represent the surface of the water. So, any positive number will be the position above water and any netative number will be the position below water. That is why, while - 4 represents the position of the dolphin below water, 4 represents the position 4 feet above water.
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Step-by-step explanation:

f^-1(1024)=(2*1024-2/5)^-1= 0.000488

f(0.000488)=2*0.000488-2/5= -0.399

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If f(x) = 7(2-1)+8, what is the value of f(11) ?
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Answer:

f ( 2 ) = 20

Explanation:

To evaluate  f ( 2 )

substitute x = 2 into  f ( x )

f ( 2 ) = ( × 2 2 − ( 4 × 2 × x = 28− 8 = 20

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3 years ago
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A rock thrown vertically upward from the surface of the moon at a velocity of 36​m/sec reaches a height of s = 36t - 0.8 t^2 met
Verdich [7]

Answer:

a. The rock's velocity is v(t)=36-1.6t \:{(m/s)}  and the acceleration is a(t)=-1.6  \:{(m/s^2)}

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

  • Velocity is defined as the rate of change of position or the rate of displacement. v(t)=\frac{ds}{dt}
  • Acceleration is defined as the rate of change of velocity. a(t)=\frac{dv}{dt}

a.

The rock's velocity is the derivative of the height function s(t) = 36t - 0.8 t^2

v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t

The rock's acceleration is the derivative of the velocity function v(t)=36-1.6t

a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6

b. The rock will reach its highest point when the velocity becomes zero.

v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = \frac{405}{2} =202.5 \:m

To find the time it reach half its maximum height, we need to solve

36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45

The rock is aloft for 45 seconds.

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Answer:

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Step-by-step explanation:

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5|2x - 7| = 20

divide both sides by 5:

|2x - 7| = 4

break into 2 equations:

2x - 7 = 4, 2x - 7 = -4

solve both equations and collect solutions:

2x = 11, 2x = 3

x = \frac{11}{2}, \frac{3}{2}

6 0
3 years ago
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