What is the probability that another student has the same birthday as you?

Tommy has a piece of toast that has butter on one side, and he dropped it twice. Both times, it landed with the butter side up.

Shkiper50 [21]

Since probability is the measure of the likelihood that an event will occur.so, maybe 5.

5 0

3 years ago

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Is it an irrational numbet

Naddik [55]

Answer:

Step-by-step explanation:

An irrational number is a number that cannot be expressed as a fraction. It has to have decimals that don't repeat or end. So your answer is no

8 0

3 years ago

14. Janie Christopher lent $6000 to a friend for 90 days at 12%. After 30 days, she sold the note to a third party for $6000. Wh

Elenna [48]

Can you do plsssssssss

help me plssssssssssss meee

3 0

2 years ago

A stack of pennies and dimes has a total value of $2.31. How many dimes are in the stack if there are twice as many dimes as pen

bixtya [17]

2 dollars in dimes i'm guessing and 31 pennies and what grade level this suppose to be

7 0

4 years ago

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Tan(A—B)=tanA—tanB/1+tanAtanB

dybincka [34]

Answer:

tan(A−B)=tan(A)−tan(B)/1+tan(A)tan(B)

1+tan(A)tan(B) To prove this, we should know that:

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)AND

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)ANDcos(A−B)=cos(A)cos(B)+sin(A)sin(B)

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)ANDcos(A−B)=cos(A)cos(B)+sin(A)sin(B)We start with:

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)ANDcos(A−B)=cos(A)cos(B)+sin(A)sin(B)We start with:tan(A−B)=sin(A−B)/cos(A−B)

cos(A−B)tan(A−B)=sin(A)/cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0
cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0
cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0
cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0The question on this page is asking for specific formula, which comes from Case 4, so I will only solve Case 4 here.

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0The question on this page is asking for specific formula, which comes from Case 4, so I will only solve Case 4 here.Dividing numerator and denominator in (1) by cos(A)cos(B) we get:

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0The question on this page is asking for specific formula, which comes from Case 4, so I will only solve Case 4 here.Dividing numerator and denominator in (1) by cos(A)cos(B) we get:tan(A−B)=sin(A)/cos(B)−cos(A)/sin(B)/;2cosA)cos(B)/cos(A)cos(B)+sin(A)sin(B)/cos(A)cos(B)

cos(A)cos(B)tan(A−B)=sin(A)cos(B)/cos(A)cos(B)−cos(A)sin(B)cos/(A)cos(B)cos/(A)cos(B)co(A)cos(B)+sin(A)sin(B)/cos(A)cos(B)

cos(A)cos(B)tan(A−B)=sin(A)/cos(A)−sin(B)cos(B)1+sin(A)cos(A)sin(B)cos(B)

cos(A)−sin(B)cos(B)1+sin(A)cos(A)sin(B)cos(B) tan(A−B)=tan(A)−tan(B)/1+tan(A)tan(B)

Step-by-step explanation:

Hope it is helpful....

3 0

3 years ago