it takes 1 gallon of diesel 1 fuel for a cruise ship to trarel 6 inches how many gallons will it take for the cruise ship to travel 1 mile

Answers

Answer 1

Answer: The last cruise I went on had a cocktail party for returning guests, which the captain attended to participate in a question and answer conversation. One of the questions asked pertained to the amount of fuel a cruise ship uses. I can't remember everything that was said, but I do remember the captain saying one gallon of fuel moved the ship a mere 17 inches!! So I would imagine close to a thousand gallons of fuel would be needed to move the ship one mile..

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Need help ASAP!!!!!!!Which pair of numbers has an LCM of 60?
a. 2 and 12
b.5 and 12
c.6 and 12
d.3 and 12

Pls answer ASAP

Answers

Answer: 5 and 12

Step-by-step explanation:

i did the quiz

Solve the equation.
1) 2(1.5n + 4) -6n= -7

2) 5s - 2 + 3 (s - 11) = 5

Answers

I hope this helps you

Can someone help me in this trig question, please? thanks A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates 5pi/12

Answers

The exact value of the position of the rider after the carousel rotates 5π/12 is 5 (-√2 + √6), 5(√2 + √6).

The position

Since the position of the carousel is (x, y) = (20cosθ, 20sinθ) and we need to find the position when θ = 5π/12 = 5π/12 × 180 = 75°

So, substituting the value of θ into the positions, we have

(20cos75°, 20sin75°)

The value of 20cos75°

20cos75° = 20cos(45 + 30)

Using the compound angle formula

cos(A + B) = cosAcosB - sinAsinB

With A = 45 and B = 30

cos(45 + 30) = cos45cos30 - sin45sin30

= 1/√2 × √3/2 - 1/√2 × 1/2

= 1/2√2(√3 - 1)

= 1/2√2(√3 - 1) × √2/√2

= √2(√3 - 1)/4

= (√6 - √2)/4

= (-√2 + √6)/4

So, 20cos75° = 20 × (-√2 + √6)/4

= 5 (-√2 + √6)

The value of 20sin75°

20sin75° = sin(45 + 30)

Using the compound angle formula

sin(A + B) = sinAcosB + cosAsinB

With A = 45 and B = 30

sin(45 + 30) = sin45cos30 + cos45sin30

= 1/√2 × √3/2 + 1/√2 × 1/2

= 1/2√2(√3 + 1)

= 1/2√2(√3 + 1) × √2/√2

= √2(√3 + 1)/4

= (√6 + √2)/4

= (√2 + √6)/4

So, 20sin75° = 20 × (√2 + √6)/4

= 5(√2 + √6)

Thus, (20cos75°, 20sin75°) = 5 (-√2 + √6), 5(√2 + √6).

So, the exact value of the position of the rider after the carousel rotates 5π/12 is 5 (-√2 + √6), 5(√2 + √6).

Learn more about position here:

brainly.com/question/11001232

$\bf \textit{the position of the rider is clearly }20cos\left( (5\pi )/(12) \right)~~,~~20sin\left( (5\pi )/(12) \right)\n\n-------------------------------\n\n\cfrac{5}{12}\implies \cfrac{2+3}{12}\implies \cfrac{2}{12}+\cfrac{3}{12}\implies \cfrac{1}{6}+\cfrac{1}{4}\n\n\n\textit{therefore then }\qquad \cfrac{5\pi }{12}\implies \cfrac{1\pi }{6}+\cfrac{1\pi }{4}\implies \cfrac{\pi }{6}+\cfrac{\pi }{4}\n\n-------------------------------$

$\bf \textit{Sum and Difference Identities}\n\nsin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)\n\ncos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)\n\n-------------------------------\n\ncos\left( (\pi )/(6)+(\pi )/(4) \right)=cos\left( (\pi )/(6)\right)cos\left((\pi )/(4) \right)-sin\left( (\pi )/(6)\right)sin\left((\pi )/(4) \right)$

$\bf cos\left( (\pi )/(6)+(\pi )/(4) \right)=\cfrac{√(3)}{2}\cdot \cfrac{√(2)}{2}-\cfrac{1}{2}\cdot \cfrac{√(2)}{2}\implies \cfrac{√(6)}{4}-\cfrac{√(2)}{4}\implies \boxed{\cfrac{√(6)-√(2)}{4}}\n\n\nsin\left( (\pi )/(6)+(\pi )/(4) \right)=sin\left( (\pi )/(6)\right)cos\left( (\pi )/(4) \right)+cos\left( (\pi )/(6)\right)sin\left((\pi )/(4) \right)$

$\bf sin\left( (\pi )/(6)+(\pi )/(4) \right)=\cfrac{1}{2}\cdot \cfrac{√(2)}{2}+\cfrac{√(3)}{2}\cdot \cfrac{√(2)}{2}\implies \cfrac{√(2)}{4}+\cfrac{√(6)}{4}\implies \boxed{\cfrac{√(2)+√(6)}{4}}\n\n-------------------------------\n\n20\left( \cfrac{√(6)-√(2)}{4} \right)\implies 5(-√(2)+√(6))\n\n\n20\left( \cfrac{√(2)+√(6)}{4} \right)\implies 5(√(2)+√(6))$

Use the formula for computing future value using compound interest to determine the value of an account at the end of 6 years if a principal amount of $6,000 is deposited in an account at an annual interest rate of 6% and the interest is compounded quarterly.

Answers

Answer: the value of the account at the end of 6 years is is $8577

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 6000

r = 6% = 6/100 = 0.06

n = 4 because it was compounded 4 times in a year.

t = 6 years

Therefore,.

A = 6000(1+0.06/4)^4 × 6

A = 6000(1+0.015)^24

A = 6000(1.015)^24

A = $8577

DJ. is on a long 18.1-mile hike through Yellowstone. if he wants to spread out his journey evenly over 5 hours, far does he need to hike how every hour?

Answers

3.62 mph because 18.1 divided by 5 = 3.62

How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

Answers

$LHS\n \n =\frac { \tan ^( 2 ){ x-\sin ^( 2 ){ x } } }{ \tan { x } } \n \n =\frac { 1 }{ \tan { x } } \left( \tan ^( 2 ){ x-\sin ^( 2 ){ x } } \right)$

$\n \n =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^( 2 ){ x } }{ \cos ^( 2 ){ x } } -\frac { \sin ^( 2 ){ x\cos ^( 2 ){ x } } }{ \cos ^( 2 ){ x } } \right) \n \n =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^( 2 ){ x-\sin ^( 2 ){ x\cos ^( 2 ){ x } } } }{ \cos ^( 2 ){ x } } \right)$

$\n \n =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^( 2 ){ x\left( 1-\cos ^( 2 ){ x } \right) } }{ \cos ^( 2 ){ x } } \n \n =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^( 2 ){ x\cdot \sin ^( 2 ){ x } } }{ \cos ^( 2 ){ x } } \n \n =\frac { \cos { x } \sin ^( 4 ){ x } }{ \sin { x\cos ^( 2 ){ x } } } \n \n =\frac { \sin ^( 3 ){ x } }{ \cos { x } }$

$\n \n =\sin ^( 2 ){ x } \cdot \frac { \sin { x } }{ \cos { x } } \n \n =\sin ^( 2 ){ x } \cdot \frac { 1 }{ \frac { \cos { x } }{ \sin { x } } } \n \n =\sin ^( 2 ){ x } \cdot \frac { 1 }{ \cot { x } } \n \n =\frac { \sin ^( 2 ){ x } }{ \cot { x } } \n \n =RHS$

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