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kupik [55]
3 years ago
11

How many radians is 60 °

Mathematics
2 answers:
Shtirlitz [24]3 years ago
5 0

Answer:

1.047

Step-by-step explanation:

60° × π/180 = 1.047rad

or

From the standard conversion factor

360∘ = 2 π r a d

we may use ratio and proportion to obtain that

60 ∘ = π 3 r a d

solniwko [45]3 years ago
5 0

Answer:

pi/3

Step-by-step explanation:

To convert from degrees to radians, we multiply by pi/180

60 degress * pi/180 =  pi/3

60 degrees is pi/3 radians

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Find the quotient of 9 divided by 5/8 .
MissTica

9 / 5/8 =

9/1 / 5/8 =

9/1 * 8/5 = 72/5 = 14 2/5

5 0
3 years ago
A motorcycle can go 50 miles using one gallon of gas. About how many gallons of gas will be used to go 150 kilometers? (Hint: 1
gayaneshka [121]

Answer:

The answer to your question is 1.875 gallons

Step-by-step explanation:

Data

distance = 50 mi

volume of gas = 1 gal

volume for 150 km = ?

Process

1.- Convert 50 mi into kilometers

                         1 mile ------------------- 1.6 km

                          x        ------------------ 150 km

                          x = (150 x 1)/1.6

                          x = 93.75 mi

2.- Calculate the gallons wasted

                          1 gallon ----------------- 50 mi

                          x            ---------------- 93.75 mi

                          x = (93.75 x 1)/50

                          x = 93.75/50

                         x = 1.875 gallons

4 0
3 years ago
Please help me to prove this!​
Ymorist [56]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π              → A + B = π - C

                                              → B + C = π - A

                                              → C + A = π - B

                                              → C = π - (B +  C)

Use Sum to Product Identity:  cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B

Use the Double Angle Identity: sin 2A = 2 sin A · cos A

Use the Cofunction Identity: cos (π/2 - A) = sin A

<u>Proof LHS → Middle:</u>

\text{LHS:}\qquad \qquad \cos \bigg(\dfrac{A}{2}\bigg)+\cos \bigg(\dfrac{B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)

\text{Sum to Product:}\qquad 2\cos \bigg(\dfrac{\frac{A}{2}+\frac{B}{2}}{2}\bigg)\cdot \cos \bigg(\dfrac{\frac{A}{2}-\frac{B}{2}}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)

\text{Given:}\qquad \quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)

\text{Sum/Difference:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)

\text{Double Angle:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{2(A+B)}{2(2)}\bigg)\\\\\\.\qquad \qquad  \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+2\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)

\text{Factor:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]

\text{Cofunction:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]

\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)

\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{\pi -C}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)

LHS = Middle \checkmark

<u>Proof Middle → RHS:</u>

\text{Middle:}\qquad 4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)\\\\\\\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)

Middle = RHS \checkmark

3 0
3 years ago
Patrick and his brother each start a savings account. Patrick begins with $200
skad [1K]

Answer:

Both brothers will have the same amount of money is their saving accounts after six months.

Step-by-step explanation:

I worked this out by finding out the first 8 months of savings for both brothers:

Patrick:               His brother:

200                     150

225                     185

250                     220

275                     253

300                     290

325                     325

375                     360

400                     395

I then looked across my results to see that at 6 months both brothers had $325.

Hope this helped.

7 0
3 years ago
Find the function of f(x)=2x-3
vazorg [7]

Sorry, But I had to put my answer as a screenshot because Brainly thought it was a link.

4 0
2 years ago
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