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3 years ago

3 and 1/13 in simplest form.

Mathematics

2 answers:

defon3 years ago

8 0

Answer:

3 1/3 is already in the simplest form. It can be written as 0.230769 in decimal form (rounded to 6 decimal places).

Have an awesome day! :D

irina1246 [14]3 years ago

3 0

Answer: it's already in simplest form

Step-by-step explanation:

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15/3 + (6,5 + 4,2) - 0,6 <br><br> Please show your work, thank you :)

kifflom [539]

Answer:

15/3 + (6.5 + 4.2) - 0.6 = 15.1

Step-by-step explanation:

You can first reduce the fraction to 5 when you multiply by 3, then calculate what's in the parenthesis. So now you have 5 + 10.7 - 0.6, then just solve from left to right. Hope this helps

3 0

3 years ago

Simplity find the quotient and the remainder of (2x2+9x+8)÷(x+2)

Nuetrik [128]

That’s what I got I didn’t know how to type it so I just took a photo

8 0

3 years ago

Determine whether y varies directly with x. If so, find the constant of variation k.

kumpel [21]

Direct variation is y=kx, inverse variation is y=k/x....

-5y=-2x divide both sides by -5

y=2.5x

So this is a direct variation with a constant of 2.5

5 0

3 years ago

for the given license plate configuration, determine how many different plates are possinle if letters and digits (a) can ba rep

Inessa05 [86]

Answer:

a) 1,757,600 different plates possible.

b) 1,404,000 different plates possible.

Step-by-step explanation:

The permutations formula is important to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

(a) can be repeated:

There are 3 letters, each one with 26 possible outcomes.

2 digits, each with 10 possible outcomes

T = 26*26*26*10*10 = 1757600

1,757,600 different plates possible.

(b) cannot be repeated:

Here the permutations formula is used.

Three letters, from a set of 26.

Two digits, from a set of 10. So

$T = P_{(26,3)}*P_{(10,2)} = \frac{26!}{(26-3)!}*\frac{10!}{(10-2)!} = 1404000$

1,404,000 different plates possible.

4 0

3 years ago

Do the rational numbers form a finite set?

zysi [14]

Yea because the rationl numbers are diffrent form as a finite set

6 0

3 years ago