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3 years ago

13

A monomial of degree 7 and a leading coefficient of -3

1 answer:

OleMash [197]3 years ago

8 0

Answer:

no clue

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B%282x%2B1%29%28x-5%29%7D%7B%28x-5%29%28x%2B4%29%5E%7B2%7D%20%7D" id

dybincka [34]

i) The given function is

$f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}$

The domain is

$(x-5)(x+4)^2\ne 0$

$(x-5)\ne0,(x+4)^2\ne 0$

$x\ne5,x\ne -4$

ii) For vertical asymptotes, we simplify the function to get;

$f(x)=\frac{(2x+1)}{(x+4)^2}$

The vertical asymptote occurs at

$(x+4)^2=0$

$x=-4$

iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

$2x+1=0$

$2x=-1$

$x=-\frac{1}{2}$

iv) To find the y-intercept, we substitute $x=0$ into the reduced fraction.

$f(0)=\frac{(2(0)+1)}{(0+4)^2}$

$f(0)=\frac{(1)}{(4)^2}$

$f(0)=\frac{1}{16}$

v) The horizontal asymptote is given by;

$lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0$

The horizontal asymptote is $y=0$ .

vi) The function has a hole at $x-5=0$ .

Thus at $x=5$ .

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

6 0

3 years ago

Fast I need help do t understand !!!!!

goblinko [34]

Answer:

....................21 .....................

6 0

3 years ago

Read 2 more answers

Find an equation of the sphere containing all surface points P = (x, y, z) such that the distance from P to A(−3, 6, 3) is "twic

Marina86 [1]

Answer:

Equation of Sphere = $x^{2}$ + $y^{2}$ + $z^{2}$ - 18x - 4/3y +10z + 142/3 = 0

Step-by-step explanation:

Data Given:

P = (x,y,z)

Distance from P to A (-3,6,3) = Twice the distance from P to B(6,2,-3)

Solution:

Find the equation of the sphere:

It is given that:

PA = 2PB

Squaring both sides:

$(PA)^{2} = (2PB)^{2}$

$(PA)^{2} = 4 (PB)^{2}$

$(x - (-3))^{2}$ + $(y - 6)^{2}$ + $(z-3)^{2}$ = 4 x $(x-6)^{2}$ + $(y-2)^{2}$ + $(z-(-3))^{2}$

Solving the above equation:

$(x + 3))^{2}$ + $(y - 6)^{2}$ + $(z-3)^{2}$ = 4 x { $(x-6)^{2}$ + $(y-2)^{2}$ + $(z + 3))^{2}$ }

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 { $x^{2}$ + 36 - 12x + $y^{2}$ + 4 - 4y + $z^{2}$ + 9 + 6z}

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 $x^{2}$ + 144 - 48x +4 $y^{2}$ + 16 - 16y + 4 $z^{2}$ + 36 + 24z

Putting the right hand side = 0 and solving the equation:

$x^{2}$ - 4 $x^{2}$ + $y^{2}$ - 4 $y^{2}$ + $z^{2}$ - 4 $z^{2}$ + 6x + 48x - 12y +16y -6z - 24z + 9 + 36 + 9 -144 - 16 - 36 = 0

-3 $x^{2}$ - 3 $y^{2}$ -3 $z^{2}$ + 54x + 4y -30z -142 = 0

Taking (-) common

- ( 3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142) = 0

3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142 = 0

dividing the whole equation by 3

$x^{2}$ + $y^{2}$ + $z^{2}$ - 54/3x - 4/3y +30/3z + 142/3 = 0

$x^{2}$ + $y^{2}$ + $z^{2}$ - 54/3x - 4/3y +30/3z + 142/3 = 0

$x^{2}$ + $y^{2}$ + $z^{2}$ - 18x - 4/3y +10z + 142/3 = 0

7 0

3 years ago

Simplify -2(-4x + 5)

Bingel [31]

Answer:

the answer would be 8x-10

3 0

3 years ago

How to do the problem I don't get it

DaniilM [7]

5 days * $125 daily fee + $??.?? Deposit = $700

5*125+x=700
625+x=700
x=75

$75 Deposit

I'm not sure what "the rate of change" is

sorry but i hope i helped

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5 0

3 years ago