All categories

3 years ago

If f(x)=3x+1 and g(x)=x^2-6 find (f-g)(x)

Mathematics

2 answers:

8090 [49]3 years ago

6 0

Hi there!

• f(x) = 3x + 1
• g(x) = x² - 6

Then,
According to th' question :-

(f - g)(x) = f(x) - g(x)

= 3x + 1 - (x² - 6)

= 3x + 1 - x² + 6

= -x² + 3x + 7

Hence,
Option 2ⁿᵈ : -x² + 3x + 7 is Correct.

~ Hope it helps!

lord [1]3 years ago

6 0

Answer:

Option B. -x² + 3x + 7

Step-by-step explanation:

The given functions are f(x) = 3x + 1 and g(x) = x² - 6

we have to find the value of (f - g)(x)

Since (f - g)(x) = f(x) - g(x)

By replacing the values of f(x) and g(x) in the equation.

f(x) - g(x) = (3x + 1) - (x² - 6)

= 3x + 1 - x² + 6

= 3x + 7 - x²

= -x² + 3x + 7

Therefore, option B. f(x) - g(x) = -x² + 3x + 7 is the correct option.

You might be interested in

A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

Immediately after launch, the rocket is subject to gravity, so its acceleration is

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$

$\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}$

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

$\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}$

and

$\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m$

$\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m$

$\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}$

b. The rocket stays in the air for as long as it takes until $z=0$ , where $z$ is the $z$ -component of the position vector.

$10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s$

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

$\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}$

c. The rocket reaches its maximum height when its vertical velocity (the $z$ -component) is 0, at which point we have

$-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)$

$\implies\boxed{z_{\rm max}=125,010\,\rm m}$

7 0

3 years ago

I need help pls I WILL GIVE BRAINLIEST Pls help

uranmaximum [27]

Answer:

1. The square root of 100

2.c=27.11

d=13.96

e=12.6

Step by step explanation: 

1. 2×50=100 the geometric mean is the square root of 100

2. The diagram is made up of three right angled triangles the big outer one and two triangles inside the big one thus we can use the pythagorean theorem to come up with expressions which will help us in solving the unknown parts as below.

c²=30²-d²

c²=e²+24²

d²=e²+6²

both the following add up to c² meaning they are equal:

c²=30²-d²

c²=e²+24²

thus

30²-d²=e²+24²

I want to remain with d² only thus;

30²-24²-e²=d²

900-576=324 (squareroot of 324=18) so

d²=18²-e² and d²=e²+6²