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For f(x)=4x+1 and g(x)=x^2-5 find (f-g)(x)

Marrrta [24]

$f(x)=4x+1,\ g(x)=x^2-5\\\\(f-g)(x)=f(x)-g(x)\\\\\text{Therefore}\\\\(f-g)(x)=(4x+1)-(x^2-5)=4x+1-x^2+5=-x^2+4x+6$

6 0

3 years ago

Read 2 more answers

PLEASE I BEG SOMEONE TO HELP

boyakko [2]

Answer:

16 + 3n

Step-by-step explanation:

a = 19 ; common difference = 22 - 19 = 3

a(n) = a + (n -1 ) *d

= 19 + (n-1)*3

= 19 + 3n - 3

= 16 + 3n

4 0

3 years ago

If f(x)=2x^3-6x^2-16x-20f(x)=2x *3 −6x *2−16x−20 and f(5)=0, then find all of the zeros of f(x)f(x) algebraically.

mihalych1998 [28]

The zeros of the cubic function f(x) = 2x³ - 6x² - 16x - 20 are given as follows:

x = 5, x = -1 + i, x = -1 - i.

<h3>How to obtain the solutions to the equation?</h3>

The equation is defined by the rule presented as follows:

f(x) = 2x³ - 6x² - 16x - 20.

One solution for the equation is given as follows:

x = 5.

Because f(5) = 0.

Then (x - 5) is a linear factor of the function f(x), which can be written as follows:

2x³ - 6x² - 16x - 20 = (ax² + bx + c)(x - 5).

This is because the product of a linear function and a quadratic function results in a cubic function.

Now we expand the right side to begin finding the coefficients of the quadratic function that we are going to solve to find the remaining zeros:

2x³ - 6x² - 16x - 20 = = ax³ + (b - 5a)x² + (c - 5b)x - 5c.

Then these coefficients are obtained comparing the left and the right side of the equality as follows:

a = 2.

-5c = -20 -> c = 4.

b = -6 + 5a = 4.

Hence the equation is:

2x² + 4x + 4.

Using a quadratic equation calculator, the remaining zeros are given as follows:

x = -1 + i.

x = -1 - i.

More can be learned about the solutions of an equation at brainly.com/question/25896797

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8 0

1 year ago

The endpoints of a line segment graphed on a Cartesian coordinate system are (4, 1) and (-2, -4). What are the coordinates of th

Masja [62]

The midpoint of (x1,y1) and (x2,y2) is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

just average them

(4,1) and (-2,4)
the midpoint is $(\frac{4-2}{2},\frac{1-4}{2})=(\frac{2}{2},\frac{-3}{2})=$ (1,-1.5)

5 0

3 years ago

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Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)

Mashcka [7]

The cone equation gives

$z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2$

which means that the intersection of the cone and sphere occurs at

$x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92$

i.e. along the vertical cylinder of radius $\dfrac3{\sqrt2}$ when $z=\dfrac3{\sqrt2}$ .

We can parameterize the spherical cap in spherical coordinates by

$\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle$

where $0\le\theta\le2\pi$ and $0\le\varphi\le\dfrac\pi4$ , which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is $\dfrac3{\sqrt2}$ . So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

$\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4$

Now the surface area of the cap is given by the surface integral,

$\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du$
$=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}$
$=18\pi\left(1-\dfrac1{\sqrt2}\right)$
$=9(2-\sqrt2)\pi$

3 0

3 years ago