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3 years ago

If f(x)=5x^3-2 and g(x)=2x+1, find (f+g)(x).

Mathematics

2 answers:

KonstantinChe [14]3 years ago

7 0

Answer:

-31

Step-by-step explanation:

SashulF [63]3 years ago

3 0

$(f+g)(x)=f(x)+g(x)\\\\
(f+g)(x)=5x^3-2+2x+1\\
(f+g)(x)=5x^3+2x-1$

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Solve the logarithmic equation. Round to the nearest ten-thousandth if necessary. log(x+9)-log x=3

zheka24 [161]

(√6,0) (√5,-1) (-√6,0) (-√5,-1)

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3 years ago

How many solutions are there to this equation 3x(x-4)+5-x=2x-7

stich3 [128]

Answer:

2 solutions

Step-by-step explanation:

I like to use a graphing calculator to find solutions for equations like these. The two solutions are ...

x = 1
x = 4

To solve this algebraically, it is convenient to subtract 2x-7 from both sides of the equation:

3x(x -4) +5 -x -(2x -7) = 0

3x^2 -12x +5 -x -2x +7 = 0 . . . . . eliminate parentheses

3x^2 -15x +12 = 0 . . . . . . . . . . . . collect terms

3(x -1)(x -4) = 0 . . . . . . . . . . . . . . . factor

The values of x that make these factors zero are x=1 and x=4. These are the solutions to the equation. There are two solutions.

<em>Alternate method</em>

Once you get to the quadratic form, you can find the number of solutions without actually finding the solutions. The discriminant is ...

d = b^2 -4ac . . . . where a, b, c are the coefficients in the form ax^2+bx+c

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4 0

3 years ago

Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En

ArbitrLikvidat [17]

Answer:

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , c) $x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$ , $y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$ .

Step-by-step explanation:

The equation of the circle is:

$x^{2} + (y-1)^{2} = 16$

After some algebraic and trigonometric handling:

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1$

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t$

Where:

$\frac{x}{4} = \cos t$

$\frac{y-1}{4} = \sin t$

Finally,

$x = 4\cdot \cos t$

$y = 1 + 4\cdot \sin t$

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

c) $x = 4\cdot \cos t''$ , $y = 1 + 4\cdot \sin t''$

Where:

$4\cdot \cos t' = 0$

$1 + 4\cdot \sin t' = 5$

The solution is $t' = \frac{\pi}{2}$

The parametric equations are:

$x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$

$y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7Ba%2Bb%7D%20%7D%7B%5Csqrt%7Ba-b%7D%20%7D%20%2B%5Cfrac%7B%5Csqrt%7Ba-b%7D%20

andrezito [222]

Answer:

$\frac{(a+b)}{a^{2} -b^{2} }$ + $\frac{(a-b)}{a^{2} -b^{2} }$ = $\frac{2a}{a^{2} -b^{2} }$

Step-by-step explanation:

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makvit [3.9K]

3947422086
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