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3 years ago

IM STRESSED OUT AND MY TEACHER ALREADY CALLED MY MOM

Mathematics

2 answers:

Hoochie [10]3 years ago

6 0

So i’m pretty sure you’re supposed to add them. first you have to find a common denominator. the easiest way to do this is to multiply the denominators! so
3•7=21
we have our denominator
now we have to convert both fractions into new fractions with the denominator
so for the first fraction, we got 21 by multiplying by 7, so we have to multiply the whole equation by 7:
2•7/3•7
that makes 14/21 (our first fraction!)
now the second fraction we multiplied by 3, so:
5•3/7•3
that makes 15/21
now we have to add them!
14/21 + 15/21 = 29/21
there’s one complete number in that fraction, because 21/21 reduces to 1/1 which is just one. but we still have 8 left over, so the simplified answer is:
1 8/21

hope this helped!!

kondaur [170]3 years ago

5 0

Answer:

B. 1 8/21

Step-by-step explanation:

trust

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erastova [34]

Answer:

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3 years ago

If the system below were written as a matrix equation, by which matrix could you multiply both sides to obtain a solution?

ludmilkaskok [199]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Below are the choices:

A.
[8 -6] 
[-5 4] 
B. 
[4 -3] 
[-2.5 2] 
C. 
[-4 5] 
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D. 
[-2 2.5] 
[3 -4]

B because it is the inverse of
(4 6) 
(5 8) 

Check 
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3 years ago

Suppose a box of Cracker Jacks contains one of 5 toy prizes: a small rubber ball, a whistle, a Captain America decoder ring, a r

lakkis [162]

Answer:

11.42 boxes

Step-by-step explanation:

For the first box bought, there is a 100% chance of getting a unique toy (since you still don't have any). E₁ = 1.

After that, there is a 4 in 5 chance of getting a unique toy from the next box, the expected number of boxes required is:

$E_2 = (\frac{4}{5})^{-1} = 1.25$

For the next unique toy, there is now a 3 in 5 chance of getting it:

$E_3 = (\frac{3}{5})^{-1} = 1.67$

Following that logic, there is a 2 in 5 chance of getting the 4th unique toy:

$E_4 = (\frac{2}{5})^{-1} = 2.5$

Finally, there is a 1 in 5 chance to get the last unique toy:

$E_5 = (\frac{1}{5})^{-1} = 5$

The expected number of boxes to obtain a full set is:

$E=E_1+E_2+E_3+E_4+E_5\\E=1+1.25+1.67+2.5+5\\E=11.42\ boxes$

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2 years ago

Find the value of x number on top is 6 and other is 9

faltersainse [42]

Answer:

x = - 12, 3

Step-by-step explanation:

By tangent secant theorem:

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