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3 years ago

What is the angle of rotation for the following figure 45° 60° 90° 120°

Mathematics

1 answer:

Tju [1.3M]3 years ago

4 0

I thinks the answer is 90 because the figure has. No sides and has an even number of triangles around it

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Calculate the unit rate: write your answer as a decimal 1/2 mile in ever 1/3 hour

Alexandra [31]

Answer:

$unit \: rate = \frac{distance}{time}$

unit rate is similar to speed / velocity:

$unit \: rate = \frac{ (\frac{1}{2} )}{ (\frac{1}{3})} \\ \\ = \frac{1}{2} \div \frac{1}{3} \\ \\ = \frac{1}{2} \times \frac{3}{1} \\ \\ = > { \boxed{ \boxed{unit \: rate = \frac{3}{2} \: miles \: per \: hour}}}$

[unit rate must have units including per <time> ]

7 0

2 years ago

QUESTION 2 84 is 30% of what number?

Tpy6a [65]

Answer: 280

Step-by-step explanation:

100/30=3.(3)

3.(3)*84=280

5 0

3 years ago

Read 2 more answers

54.00 for tickets write a unit rate for the situation

schepotkina [342]

54 dollars/1 ticket.
sorry couldnt find fraction bar used / instead .

6 0

3 years ago

(85 x 255) / 289 + 15 - 70 =

ANEK [815]

I do believe its twenty

5 0

3 years ago

Read 2 more answers

Question 1: The water level in a tank can be modeled by the function h(t)=4cos(+)+10, where t is the number of hours

Amanda [17]

Answer:

Question 1: The hours that will pass between two consecutive times, when the water is at its maximum height is π hours

Question 2: Sin of the angle is -0.8

Step-by-step explanation:

Question 1: Here we have h(t) = 4·cos(t) + 10

The maximum water level can be found by differentiating h(t) and equating the result to zero as follows;

$\frac{\mathrm{d} h(t)}{\mathrm{d} t} = \frac{\mathrm{d} \left (4cos(t) + 10 \right )}{\mathrm{d} t} = 0$

$\frac{\mathrm{d} h(t)}{\mathrm{d} t} = - 4 \times sin(t) = 0$

∴ sin(t) = 0

t = 0, π, 2π

Therefore, the hours that will pass between two consecutive times, when the water is at its maximum height = π hours.

Question 2:

B = (3, -4)

Equation of circle = x² + y² = 25

Here we have

Distance moved along x coordinate = 3

Distance moved along y coordinate = -4

Therefore, we have;

$Tan \theta = \frac{Distance \ moved \ along \ y \ coordinate}{Distance \ moved \ along \ x \ coordinate} = \frac{-4}{3}$

$\therefore \theta = Tan^{-1}(\frac{-4}{3}) = -53.13^{\circ}$

Sinθ = sin(-53.13) = -0.799≈ -0.8.

5 0

3 years ago