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3 years ago

How do you figure out whether a fraction will be a terminating decimal or a repeating decimal?

1 answer:

4 0

If<span> the prime factorization of the denominator of a fraction has only factors of 2 and factors of 5, then it can be written as something over a power of ten. This means that the decimal will terminate.</span>

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what are the steps required to determine the equation of a quadratic function given its zeros and a point?

mezya [45]

Answer:

Below

Step-by-step explanation:

The quadratic equations form is:

● ax^2+bx+c

Using the zeroes, we can write a factored form.

● a (x-x') (x-x")

x and x' are the zeroes

■■■■■■■■■■■■■■■■■■■■■■■■■■

●y = a (x-x') (x-x")

x' and x" are khown but a is not.

We are given a point so replace x and y with its coordinates to find a.

So the steps are:

● 1) Write the factored form of the quadratic equation

● 2) replace x' and x" with their values.

● 3) replace x and y with the coordinates of a khwon point.

● 4) solve the equation for a.

8 0

3 years ago

What is 365/517 of 89350?

shusha [124]

0.0000079 is your answer, I believe.

6 0

3 years ago

At a school with 100? students, 33 were taking? Arabic, 32 ?Bulgarian, and 40 Chinese. 9 students take only? Arabic, 12 take onl

Stels [109]

Answer: The answer is 4 and 32.

Step-by-step explanation: Let "A", "B" and "C" represents the set of students who were taking Arabic, Bulgarian and Chinese respectively.

The, according to the given information, we have

$n(A)=33,~~n(B)=32,~~n(C)=40,~~n(A\cap B)=14.$

Let 'p' represents the number of students who take all the three languages, then

$n(A\cap B\cap C)=p.$

Also,

$n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=9\\\\\Rightarrow n(A\cap B)+n(A\cap C)=24+p~~~~~~~~~~~~~~(a),\\\\n(A\cap B)+n(B\cap C)=20+p~~~~~~~~~~~~~~(b),\\\\n(B\cap C)+n(A\cap C)=20+p~~~~~~~~~~~~~~~(c).$

From here, we get after subtracting equation(c) from (b) that

$n(A\cap B)=n(A\cap C)=14.$

Therefore,

$p=14+14-24=4,$ and from equation (a), we find

$n(B\cap C)=24-14=10.$

Thus,

$n(A\cap B\cap C)=4$ and

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cap B\cap C)=33+32+40-9-12-20+4\\\\\Rightarrow n(A\cap B\cap C)=68.$