What is 0.03 as a fraction

Answers

Answer 1

Answer: 0.03 as a fraction is 3/100

Answer 2

Answer: $\boxed{\bold{0.03= (3)/(100) }}$

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You work for a small business that sells bicycles, tricycles, and tandem bicycles (bicycles built for two). Bicycles have one seat, one front-steering handlebars, two pedals , and two wheels. Tricycles have one seat, one front-steering handlebars, two pedals, and three wheels. Tandem Bicycles have two seats, one front-steering handlebars, four pedals, and two wheels. Part A: On Monday, you counted forty-eight tricycle wheels. How many tricycles were in the shop? Write an algebraic equation that shows the relationship between the number of wheels (w) and the number of tricycles (t). Part B: On Wednesday, there were no tandem bicycles in the shop. There were only bicycles and tricycles. There were a total of twenty-four seats and sixty-one wheels in the shop. How many bicycles and how many tricycles were in the shop? Solve algebraically and show all your work. Let a = the number of tricycles Let b = the number of bicycles Part C: A month later, there were a different number of bicycles, tricycles, and tandem bicycles in the shop. There were a total of 144 front-steering handlebars, 378 pedals, and 320 wheels. How many bicycles, tricycles, and tandem bicycles were in the shop? Solve algebraically and show all your work. Let a = the number of tricycles Let b = the number of bicycles Let c = the number of tandem bicycles.

Answers

Part A: each tricycle has three wheels, so with 48 wheels the number of tricycles was a =48/3=16 tricycles.
t=w/3 (the number of tricycles is the number of wheels divided by 3)

Part B:
The number of seats:
24=b+a (so b=24-a)
The number of seats is the sum of one seat per bicycle and one seat per a tricycle

also, 61=2a+3b (the number of wheels)

So we have:
24=b+a
b=24-a
We can substitute this for b:

61=2a+3(24-a)

and solve:
61=2a+3*24-3a
61=72-a
a=72-61
a=11

There were 11 bicycles!!
and there were 24-11 tricycles, so 13 tricycles.

Part C: each of the bikes has only one front-steering handlebar, so there were a total of 144 vehicles:

a+b+c=144

There were 378 pedals. And the number of pedals is:
2a+2b+4c=378 (the numbers 2,2,4 represent the number of pedals per vehicle)

divide by 2:
a+b+2c=189

Now, we have
a+b+2c=189
and

a+b+c=144
and we can subtract them from each other:
a+b+c-(a+b+2c)=144-189
-c=45
c=45, so there were 45 tandem bicycles!
(this also means that a+b=144-45, that is a+b=99)
now the wheels:
3a+2b+2c=320
Let's substitute c:
3a+2b+90=320

which is
3a+2b=240
We also know that a+b=99, so we can substract this from this equation:
3a+2b+-a-b=240-99
2a+b=141

and again:
2a+b-a-b=141-99
a=42 - there were 42 trycicles!!!

And the bicycles were the rest:
99-42=57 bycicles

How to change 50/12 to a mixed number?

Answers

A mixed number is one that has an integer along with a proper fraction, so that would be (for this improper fraction):

50/12 = 4 1/6

12 goes into 50, 4 times with a remainder of 2

you get: 4 and 2/12, but 2/12 can be reduced further to 1/6

so that's how i got the answer of 4 1/6

Tomato ehjhhjrhu huwdhajkwh

Answers

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Answer:

its a fruit

Step-by-step explanation:

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10(23+7)=100ℓ+100ℓ What is the value of \large\ellℓell in the equation above?

Answers

Answer:

ℓ = 1.5

Step-by-step explanation:

An equation 10(23+7)=100ℓ+100ℓ is given and we are asked to find the value of ℓ.

To solve the equation, we have to move all the values containing ℓ on one side and other values on other side of equality.

10(23+7)=100ℓ+100ℓ

Multiply 10 with 23 and 7 which gives

(10 * 23 ) + ( 10 * 7) = 200 ℓ

Solving the above equation,

230 + 70 = 200 ℓ

300 = 200 ℓ

=> ℓ = 300 / 200

=> ℓ = 3/2

=> ℓ = 1.5

What is the equation in point-slope form of the line that passes through the point (5,-5) and is parallel to the line represented by y = -8x - 2?

Answers

Answer: y + 5 = -8 ( x - 5 )

Step-by-step explanation:

If A, B,C are the angles of a triangle then prove: (the following in picture)Please help me to prove this.

Answers

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → A + B = π - C

→ C = π - (A + B)

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

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

Use the Double Angle Identity: cos 2A = 1 - 2 sin² A

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

Proof LHS → RHS:

LHS: cos A + cos B + cos C

= (cos A + cos B) + cos C

$\text{Sum to Product:}\qquad 2\cos \bigg((A+B)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Given:}\qquad 2\cos \bigg((\pi -C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C\n\n\n.\qquad \qquad =2\cos \bigg((\pi)/(2) -(C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Cofunction:}\qquad 2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Double Angle:}\qquad 2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos\bigg(2\cdot (C)/(2)\bigg)\n\n\n.\qquad \qquad \qquad =2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+1-2\sin^2 \bigg((C)/(2)\bigg)\n\n\n.\qquad \qquad \qquad =1+2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)-2\sin^2\bigg((C)/(2)\bigg)$

$\text{Factor:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((C)/(2)\bigg)\bigg]$

$\text{Given:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((\pi-(A+B))/(2)\bigg)\bigg]\n\n\n.\qquad \qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((\pi)/(2)-(A+B)/(2)\bigg)\bigg]$

$\text{Cofunction:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\cos\bigg((A+B)/(2)\bigg)\bigg]$

$\text{Product to Sum:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[2\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\n\n\n.\qquad \qquad \qquad \qquad =1+4\sin \bigg((C)/(2)\bigg)\bigg[\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\n\n\n.\qquad \qquad \qquad \qquad =1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)$

$\text{LHS = RHS:}\ 1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)=1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)\quad \checkmark$

The proof for this is simple. Let's say that A + B + C = π. From here on we require several trigonometric identities that must be applied.

$\cos \left(A\right)+\cos \left(B\right)+\cos \left(C\right) \n= 2 * cos((A + B) / 2) * cos((A - B) / 2) + \cos C \n= 2 * cos((\pi /2) - (C/2)) * cos((A - B) / 2) +\cos C \n= 2 * sin(C/2) * cos((A - B) / 2) + (1 - 2 * sin^2 (C/2)) \n= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin (C/2) \n= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin((\pi /2) - (A + B)/2 ))\n= 1 + 2 sin (C/2) * cos((A - B) / 2) - cos((A + B)/ 2)\n= 1 + 2 sin (C/2) * 2 sin (A/2) * sin(B/2) \n= 1 + 4 sin(A/2) sin(B/2) sin(C/2)$

Hope that helps!

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