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3 years ago

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Shawn and dorian rented bikes from two different rental shops. the prices in dollars, y, of renting bikes from the two different

shops for x hours is shown.shop shawn used: y = 10 3.5xshop dorian used: y = 6xif shawn and dorian each rented bikes for the same number of hours and each paid the same price, how much did each pay for the rental?$

2 answers:

Hitman42 [59]3 years ago

6 0

They paid $24 each for the rental.

Hoochie [10]3 years ago

4 0

Answer:

option d

Step-by-step explanation:

i know things...

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Please help! :)

sweet-ann [11.9K]

Answer:

Second option is correct

$Arc\ length = \frac{3}{4} \pi\ in$

Step-by-step explanation:

Given:

Central angle = 15°

Radius of the circle = 9 in

Arc length = ?

Given formula is

$\frac{Arc\ length}{Circumfrance} = \frac{n}{360}$

Where n is the central angle of the sector.

Write the given formula for Arc length of the sector.

$Arc\ length = Circumfrance\times \frac{n}{360}$

We know that the the circumference of the circle is $2\pi r$ , where r is the radius of the circle,

$Arc\ length = 2\pi r\times \frac{n}{360}$

Now we substitute central angle value and radius value in above equation.

$Arc\ length = 2\pi 9\times \frac{15}{360}$

$Arc\ length = 18\times \pi \frac{15}{360}$

$Arc\ length = \frac{18\times 15}{360} \pi$ ----------( $\frac{18}{360}=\frac{1}{20}$ )

$Arc\ length = \frac{15}{20} \pi$

Divide the numerator and denominator by 5.

$Arc\ length = \frac{3}{4} \pi\ in$

Therefore the arc length of the sector is $\frac{3}{4} \pi\ in$

4 0

3 years ago

How to find vertical and horizontal asymptotes of polynomials?

guajiro [1.7K]

Vertical asymptotes are when y tends to infinity, which in other words means you need a fraction where you're dividing by zero. So to find the vertical asymptotes, solve the denominator of the fraction and the solutions are the asymptotes since that is where it will be dividing by zero.

Horizontal asymptotes are the opposite and is where x tends to infinity -- This can be done by dividing through with the xs still there, since you are effectively just looking at multiples of infinity and infinity squared and so on. When you have an x to the power you can forget about all the other x's on the same half of the fraction, since infinity squared is infinitely bigger than infinity - yes it's abstract and a bit stupid I know.
If the degree of the polynomial on the denominator is larger than the numerator, there is an asymptote at y=0 (since the bottom becomes infinitely larger than the top). If it is lower than the numerator, you have to divide through by long division and you end up with an oblique asymptote i.e. an asymptote where the dotted line would follow a linear equation e.g. y=x

For example
$f(x)= \frac{6x^2-3x+4}{2x^2-8}$

The vertical asymptote can be found by solving 2x^2-8=0 i.e. x=+-2 so our vertical asymptotes are at 2 and -2.

The horizontal asymptote, think of all the x's as infinity. SInce there are x^2s, we dont need to worry about the numbers or the other x's
Therefore we are left with 6x^2/2x^2, which gives us 3. Therefore our horizontal asymptote would be at y=3.

7 0

3 years ago

At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given

Nostrana [21]

Answer:

a. The initial amount was 10 mg.

b. The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.

c. The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.

d. The time until only 1 mg of the drug remains in the body is 11.6 hours.

Step-by-step explanation:

You know that at time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by :

$A=10*(0.82)^{t}$

a. The initial quantity occurs when time t is the initial t, that is, t is equal to 0. Then:

$A=10*(0.82)^{t}=10*(0.82)^{0}=10*1\\$

A=10

<u><em>The initial amount was 10 mg.</em></u>

b. Considering that an exponential growth is determined by:

$A=A0*(1-r)^{t}$ , where A is the amount after a certain number of a certain time, Ao is the initial amount, r is the rate and t is the time, so the percentage of the drug that leaves the body each hour is :

1-r=0.82

Solving:

1-r -1= 0.82 -1

-r= -0.18

r= 0.18

<u><em>The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.</em></u>

c. The amount of drug that remains in the body 6 hours after dosing is when t = 6:

$A=10*(0.82)^{6}$

Solving:

A= 3.04 mg

<u><em>The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.</em></u>

d. The time that passes until only 1 mg of the drug remains in the body is calculated taking into account that A = 1 mg:

$1=10*(0.82)^{t}$

Solving:

$0.1=(0.82)^{t}$

㏒ 0.1= t*㏒ 0.82

㏒ 0.1 ÷ ㏒ 0.82= t

11.6 hours= t

<u><em> The time until only 1 mg of the drug remains in the body is 11.6 hours.</em></u>

4 0

3 years ago

Pls help me ..................

valina [46]

Answer:

B.

D.

E.

Step-by-step explanation:

7 0

3 years ago

Write the product in its simplest form 3w^5×7w^5

harkovskaia [24]

Answer:

21 $w^{10}$

Step-by-step explanation:

Using the rule of exponents

$a^{m}$ × $a^{n}$ ⇔ $a^{(m+n)}$

Given

3 $w^{5}$ × 7 $w^{5}$

= 3 × 7 × $w^{(5+5)}$

= 21 $w^{10}$

5 0

3 years ago