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katovenus [111]
3 years ago
13

Please answer the following essay question:

Mathematics
2 answers:
Masteriza [31]3 years ago
5 0

Answer: The denominator of a fraction tells you the relative size of the pieces. Therefore, the reason fractions need a common denominator before adding or subtracting is so that the numbers of pieces you are adding/subtracting are all the same size.

ki77a [65]3 years ago
5 0

Answer:

We need to have a common denominator in order to subtract different fractions because if we have an example like.

5/2 - 6/8 we cannot do that because the denominator is not alike.

We have to use multiplication to help us here.

2 x 8 = 16 !

So. no we have 5/16

so now we do 5 x 8 = 40

So now my new fraction is - 40/16 BUTTTT 40/16 is an improper fraction. so 40/16 = 5/2

Now we are back at the square 1 so now we have to fix 6/8.

8 x 6 = 48

6 x 6 = 36.

--

2 x 24 = 48.

5 x 24 = 120 or 12

now we have 12/48 and  36/48.

We can subtract now!

Hope this helped for problems like this.

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What is the measure of x
Dafna1 [17]

Answer:

58 degrees

Step-by-step explanation:

XYZ is 90 degrees. XYO is 32 degrees. Angle x takes up the space leftover in the right angle, so simply subtract 32 from 90.

90-32= 58

Hope this help! (if it does pls consider giving me brainliest)

7 0
2 years ago
Solve <br> 4x+6&lt;-6<br><br> I could really use the help
aleksandr82 [10.1K]

Answer:

x < -3

Step-by-step explanation:

First subtract 6 from both sides of the inequality.

4x+6-6

4x

Divide 4 on both sides.

\frac{4x}{4} < \frac{-12}{4}

x

5 0
3 years ago
3100 times 43 how much is this
Soloha48 [4]

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133,300

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3100 x 43 = 133,300

7 0
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Read 2 more answers
Find two rational expressions a / b and c / d that produce the result x − 1 / x2 when using the following operations. Answers
Mars2501 [29]

Answer:

a) Let \frac{a}{b}=\frac{-1}{x^2}, \text{ and } \frac{c}{d}=\frac{1}{x}.

Observe that

\frac{a}{b}+\frac{c}{d}=\frac{-1}{x^2}+\frac{1}{x}=\frac{-x+x^2}{x^3}=\frac{x(x-1)}{xx^2}=\frac{x-1}{x^2}

b)

Let \frac{a}{b}=\frac{1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x^2}.

Observe that

\frac{a}{b}-\frac{c}{d}=\frac{1}{x}-\frac{1}{x^2}=\frac{x^2-x}{x^3}=\frac{x(x-1)}{xx^2}=\frac{x-1}{x^2}

c)

Let \frac{a}{b}=\frac{x-1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x}.

Observe that

\frac{a}{b}*\frac{c}{d}=\frac{x-1}{x}*\frac{1}{x}=\frac{(x-1)1}{x*x}=\frac{x-1}{x^2}

d)

Let \frac{a}{b}=\frac{x-1}{x}, \text{ and } \frac{c}{d}=\frac{x}{1}.

Observe that

\frac{a}{b}\div\frac{c}{d}=\frac{x-1}{x}\div\frac{x}{1}=\frac{x-1}{x}*\frac{1}{x}=\frac{x-1}{x^2}

3 0
3 years ago
2) y - 1+x<br> -x - 3+ y
Art [367]

wait.

y - 1+x-x - 3+ y . . .or. . . y - 1+x . . . -x - 3+ y

5 0
3 years ago
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