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2 years ago

Given the system below:

Mathematics

2 answers:

Norma-Jean [14]2 years ago

3 0

$f(x)$ and $g(x)$ are literally the same function: their output is twice the input.

So, no matter which input you feed to f and g, they will always return the same output.

All of the option apply, as well as every other possible number.

hram777 [196]2 years ago

3 0

Answer:

B) x = 1

C) x = 4

Step-by-step explanation:

Given:

f(x) = 2x

g(x) = 2^x

We have to find at what value of x, the above two functions are equal in value.

Now we have to check with the each options.

When x = 0, f(0) = 2(0) = 0 ; g(0) = 2^0 = 1 [Anything power 0, is 1]

When x = 1, f(1) = 2(1) = 2 ; g(1) = 2^1 = 2.

when x = 2, f(2) = 2(2) = 4; g(2) = 2^2 = 4

when x = 3, f(3) = 2(3) = 6, g(3) = 2^3 = 8

when x = 4, f(4) = 2(4) = 8, g(4) = 2^4 = 16

The two functions are equal in value when x = 1 and x = 2.

You might be interested in

State the domain and range of the following relation<br> x^2+y^2=16

Vlada [557]

The domain and range is [-4, 4] and [0, 4]

<h3 /><h3>What is Domain and range?</h3>

The domain of a function is the set of values that we are allowed to plug into our function.

The range of a function is the set of values that the function assumes.

x² + y² = 16

y = √16 - x²

For domain under root should not be negative quantity,

16 - x²≥0

16≥x²

So, x≤4 or x≥4

Thus, the domain is [-4, 4]

Range:

y is maximum at x=0, y=4

y is minimum at x=4, y=0

Thus, range = [0, 4]

Learn more about domain and range here:

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8 0

1 year ago

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .