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igomit [66]
2 years ago
10

You travel 5 miles downstream in a kayak at a speed of r miles per hour. You turn around and travel 6 miles upstream at a speed

of 0.6r miles per hour. Finally, you turn around and return to your original starting point at a speed of r+1 miles per hour.
Mathematics
1 answer:
Vlad1618 [11]2 years ago
3 0
Kyle, d = rt, right? This means that t = d/r. If we call the rate of the current c, then her rate upstream is hindered by the current and is 3 - c. Downstream is 3 + c. The total time is 6 hours... 5/(3 - c) + 5/(3 + c) = 6 Once we put everything over the same denominator, we don't need it anymore. 5(3 + c) + 5(3 - c) = 6(3 - c)(3 + c) 15 + 5c + 15 - 5c = 54 - 6c2
30 = 54 - 6c26c2 = 24c2 = 4c = 2 2 mph 
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monitta
Honestly im not even sure lol
3 0
1 year ago
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Determine which equations below, when combined with the equation 3x-4y=2, will form a system with no solutions. Choose all that
svetoff [14.1K]
Two equations will not have solution if they are parallel and have different y-intercepts. Parallel lines have the same slope. In a slope-intercept form, the equation of the line can be expressed as,       
 
        y = mx + b 

where m is slope and b is the y-intercept.

Given: 3x - 4y = 2
Slope-intercept: y = 3x/4 - 1/2

A. 2y = 1.5x - 2
Slope-intercept: y = 3x/4 - 1

B. 2y = 1.5x - 1
Slope-intercept:  y = 3x/4 - 1/2

C. 3x + 4y = 2
Slope-intercept:  y = -3x/4 + 1/2

D. -4y + 3x = -2
Slope-intercept: y = 3x/4 + 1/2

Hence, the answers to this item are A and D. 
6 0
2 years ago
a shopkeeper sold goods for rs 2400 and made a profit of 25% in the process. find his profit per cent if he had sold his goods f
miv72 [106K]

case 1,

Let the CP be ₹x,

SP = ₹2400

Profit = SP – CP

= 2400 – x

Profit % = {(2400–x)/ x} × 100%

According to the question,

{(2400–x)/ x} × 100 = 25

=> (2400–x)/ x= 25 /100

=> 100(2400–x) = 25x [ cross multiplication]

=> 240000 – 100x = 25x

=> 240000 = 25x + 100x

=> 240000 = 125x

=> 240000/125 = x

=> x = 1920

So, CP = ₹1920

case 2,

SP = ₹2040

Profit = SP – CP

= 2040 – 1920

= ₹120

profit % = 120/1920 × 100%

= 16%

<h3>Thus, his profit would be 16% if he had sold his goods for ₹2040.</h3>
6 0
1 year ago
On a certain​ route, an airline carries 7000 passengers per​ month, each paying ​$30. A market survey indicates that for each​ $
KengaRu [80]

Answer:

The ticket price that maximizes revenue is $50.

The maximum monthly revenue is $250,000.

Step-by-step explanation:

We have to write a function that describes the revenue of the airline.

We know one point of this function: when the price is $30, the amount of passengers is 7000.

We also know that for an increase of $1 in the ticket price, the amount of passengers will decrease by 100.

Then, we can write the revenue as the multiplication of price and passengers:

R=p\cdot N=(30+x)(7000-x)

where x is the variation in the price of the ticket.

Then, if we derive R in function of x, and equal to 0, we will have the value of x that maximizes the revenue.

R(x)=(30+x)(7000-100x)=30\cdot7000-30\cdot100x+7000x-100x^2\\\\R(x)=-100x^2+(7000-3000)x+210000\\\\R(x)=-100x^2+4000x+210000\\\\\\\dfrac{dR}{dx}=100(-2x)+4000=0\\\\\\200x=4000\\\\x=4000/200=20

We know that the increment in price (from the $30 level) that maximizes the revenue is $20, so the price should be:

p=30+x=30+20=50

The maximum monthly revenue is:

R(x)=(30+x)(7000-100x)\\\\R(20)=(30+20)(7000-100\cdot20)\\\\R(20)=50\cdot5000\\\\R(20)=250000

3 0
2 years ago
The diagram shows the cross section of a cylindrical pipe with water lying in the bottom.
disa [49]
Hey 

So my brother posted this on Yahoo 
Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth. A right-angled triangle is formed. Length of side to the water-surface is 5 cm, the hypot is 7 cm. 

<span>What you do now is the following: </span>

<span>Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7) </span>

<span>So θ is approx 44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8° </span>

<span>The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram. </span>

<span>Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8° </span>
<span>= 88.8/360*π*7² - 24.5*sin 88.8° </span>
<span>≃ 13.5 cm² </span>
<span>(using area of ∆ = ½.a.b.sin C for the triangle) </span>



<span>b) </span>

<span>volume of water = cross-sectional area * length </span>
<span>≃ 13.5 * 30 cm³ </span>
<span>≃ 404 cm³</span>

Hoped it Helped
5 0
1 year ago
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