13 units find area of sector EFG.

Which graph is an example of a function whose parent graph is of the form y = √x?

Maksim231197 [3]

Answer:

The correct option is 1.

Step-by-step explanation:

The given parent function is

$y=\sqrt{x}$

1. Domain of the function is all positive real number including 0.

2. Range of the function is all positive real number including 0.

3. It is an increasing function. It increases at decreasing rate.

First graph increase at decreasing rate and it starts from (-4,-1), therefore the required function is

$y=\sqrt{x+4}-1$

Therefore graph 1 is an example of a function whose parent graph is of the form y = √x.

Second graph is a parabola, so it is the graph of a quadratic function.

Third graph is a rectangular hyperbola, so it is the graph of a rational functions.

Fourth graph is increasing at increasing rate, so it is the graph of an exponential function.

Therefore options 2, 3 and 4 are incorrect.

6 0

3 years ago

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The minute hand of a clock turns 6° per minute. In fifteen minutes the

Katena32 [7]

Answer:

25%

Step-by-step explanation:

15 times 6 is 90

90 is 25% of 360

8 0

3 years ago

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Mr. franklin paid $33.20 for 8 gallons of gas. What is the price of 1 gallon ?

evablogger [386]

The price of 1 gallon is $4.15. You figure this out by dividing 33.20 by 8.
33.20/8 = 4.15

3 0

3 years ago

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A rock thrown vertically upward from the surface of the moon at a velocity of 36m/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.

5 0

3 years ago

Convert a distance of 42 feet into yards.

dybincka [34]

Answer:

14 Yards

Step-by-step explanation:

8 0

3 years ago