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4 years ago

PLEASE HELP SIMPLE MATH PROBLEM WILL MARK BRAINLIEST

Mathematics

1 answer:

olga2289 [7]4 years ago

8 0

Answer:

not simple math but ok-

6,8

Step-by-step explanation:

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The picture below shows a portion of a river dam:

NNADVOKAT [17]

Hello,

If you are starting with the angle of 40 degrees, you have an adjacent side and hypotenuse. Therefore, it is a cosine relationship.

We can write:

cos(40) = adj / hyp
cos(40) = x / 220

To solve for x, simply multiply 220 by the cos(40).
You can write it as: 220 cos(40)

Good luck,
MrEQ

8 0

3 years ago

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Write the relation as a set of ordered pairs.

AveGali [126]

Answer:

Is B (-2,-1),(0,0),(2,1)

Step-by-step explanation:

x y

-2 -1

0 0

2 1

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3 years ago

1.Shelly lives 500 miles away.Paul drove 65 mph for 4 hours.How many more miles will it take for him to arrive at Shelly’s house

Sladkaya [172]

1.240 miles
65*4=260
500-260

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3 years ago

A bag contains five red marbles, four blue marbles, and three yellow marbles. You randomly pick a marble. What is the probabilit

seraphim [82]

The ratio is 9/12 and you can write it as a proportion. so you multiply 9 by 100 then divide it by 12. answer is 75% chance

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3 years ago

A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between

sattari [20]

Answer:

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) 0.09

Step-by-step explanation:

We are given the following in the question:

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)$

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) rate of increase after one day.

We put t = 1

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09$

The rate of increase after 1 day is 0.09

8 0

3 years ago