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3 years ago

How do you get horizontal asymptopes?

1 answer:

4 0

<span>1) If the degree (the largest exponent) of the denominator is bigger than the degree of the numerator, the horizontal asymptote is the x-axis (y = 0).
<span>
2) If the degree of the numerator is bigger than the denominator, there is no horizontal asymptote.</span></span>

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Solve for x 3x=36-6x

tatyana61 [14]

The correct answer is

X= 4

Please give brainliest :))

4 0

2 years ago

They are arranged randomly because manual typewriters tended to jam if the user typed too fast - therefore the arrangement was i

solmaris [256]

Answer:

Huh

Step-by-step explanation:

Huh......................

8 0

2 years ago

Read 2 more answers

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

Write an equation of a line in slope-intercept from with the given slope, slope:–3/4, y-intercept: –2

larisa [96]

Answer: y=-3/4x-2

Step-by-step explanation:

Write an equation of a line in slope-intercept from with the given slope, slope:–3/4, y-intercept: –2

The y-intercept is at (0,-2)

Then we write the equation using y=mx+b becuase that is the point slope form.

m=-3/4 which was given in the question

because y-intercept is at (0,-2)

y=-2

x=0

-2=-3/4(0)+b

-2=0+b

-2=b

now we know that m is -3/4 and b is -2

Hence the equation is y=-3/4x-2

7 0

3 years ago

It's time for another financial calculator problem. A UCF student (who has not taken FIN 2100) decides that he really needs a la

AysviL [449]

Answer:

interest rate is 38.68 %

Step-by-step explanation:

Given data

installment = $60

time = 36 months = 36/12 = 3 years

principal = $1000

to find out

interest rate

Solution

we know student pay $60 for 36 months

so he pay total = 60 × 36 = 2160

total amount pay by student = $ 2160

so we can find interest rate by given formula

rate = (1/time)(amount/Principal - 1)

put the value time amount and principal here

rate = (1/3)(2160/1000 - 1)

rate = 0.386667

interest rate is 38.68 %

5 0

3 years ago