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2 years ago

Use the identities for the cosine of a sum or a difference to write the expression as a single function of x.

Mathematics

1 answer:

lions [1.4K]2 years ago

3 0

Answer:

reeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Step-by-step explanation:

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Sally went 3.5 miles per hour on a treadmill for 5 hours. She was sore and decided to take 400 mg of Advil. The half portion sta

xenn [34]

Answer:

8 hours

Step-by-step explanation:

it says: The half portion stays in her body for 8 hours.

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3 years ago

A ship sails north for 2 hours, traveling a total of 15 miles. Then the ship turns south and travels 6 miles in an hour.

ch4aika [34]

Answer:

3 miles per hour

Step-by-step explanation:

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2 years ago

In a bag full of small marbles, 1/4 of the marbles are green 1/8 of the marbles are blue and 1/12 of the marbles are

Alik [6]

Step-by-step explanation:

white=26

1/4+1/8+1/12=6/24+3/24+2/24=11/24=(green,blue and yellow marbles)

24-11=13/24=white

26/13=2=1/24

blue=3/24

2.3=6

answer 6

8 0

2 years ago

A city has an elevation of 465 feet. Another city has an elevation of -45 feet. Which expression can be used to find the differe

KengaRu [80]

465 -(-45) = x

This creates a double negative which becomes a positive and you solution would be 465+45 =X, X = 510

6 0

2 years ago

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

8 0

2 years ago