I know...another one

The graphs below are both quadratic functions. The equation of the red graph is

Natali5045456 [20]

Answer:

D. $g(x)=\frac{1}{5}x^2$

Step-by-step explanation:

The equation of the red quadratic graph is $f(x)=x^2$ .
This is the parent quadratic function or the base of the quadratic functions.
This red quadratic graph has been dilated by a certain scale factor to obtain the blue graph. Recall that, when the scale factor is a fraction, such that $0\:$ the graph is stretched horizontally.
The dilated graph then becomes wider than the parent function.
From the given function equations, the possible value of k can only be $k=\frac{1}{5}$ .
Therefore the blue graph has equation: $g(x)=\frac{1}{5}x^2$
The correct answer is D.

6 0

3 years ago

If A has the coordinate -10 and C has the coordinate 14 and B is the midpoint of a AC, then what is the coordinates of D which i

salantis [7]

Answer:

8

Step-by-step explanation:

The midpoint of AC is ...

B = (A +C)/2 = (-10 +14)/2 = 2

The midpoint of BC is ...

D = (B +C)/2 = (2 +14)/2 = 8

The midpoint of BC is 8.

5 0

2 years ago

Pls help how do I find the variables? (offering 30 pts, don't answer unless you know the answer.)

meriva

Answer:

Since the above triangle is Iscocleles we can make our equation-

5x - 8 = 2x + 7

5x - 2x = 8 + 7

3x = 15

x = 5

Hope it helps

3 0

2 years ago

5/10 reduced to its lowest fraction

icang [17]

5/10 in the lowest fraction is 1/5

The decimal form is 0.5 if you need it

You should use the website Math wa* it helps you will a lot with math only but if you have to pay to see how the computer shows it work and steps to the answer but the answer is free just put the question and you get the answer. It is never wrong I've been using it for 4 years.

*=y so, the * is y at the end of the word

5 0

3 years ago

Read 2 more answers

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