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3 years ago

How do you add 3 fractions of different denominators

Mathematics

2 answers:

dimulka [17.4K]3 years ago

5 0

Step 1: Find the Lowest Common Multiple between the denominators.

Step 2: Multiply the numerator and denominator of each fraction by a number (the one that will get them to the lcm) so that they have the LCM as their new denominator.

Step 3: Add or subtract the numerators and keep the denominator the same.

FrozenT [24]3 years ago

3 0

First, then the denominations the same! So, find a product (which is a answer to a multiplication problem) that both numbers can go in! then change the numerator also, by the same multiplication, and then u add! Remember that before u add fractions, u always need to have same denominators!

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OLEGan [10]

Answer:

Step-by-step explanation:

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2 years ago

The product of the page numbers on two facing pages of a book is 156. Find the page numbers

Gnoma [55]

Let x be the page number then x+1 would be the other page number (because the pages are two facing pages which means they follow on consecutively)

Then

(x)(x+1) = 156

x^2 + x - 156 = 0

x = 12 or x = -13( but x can't be a negative number)

so x = 12

and the next page is 13

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3 years ago

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aliina [53]

It is true, how??Here is explanation:

Consider a quadrilateral ABCD .Join diagnol AC so two triangles ABC & ACD will form.
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3 years ago

4x − 1 < 11 solve for x

kvv77 [185]

Answer:

$\huge \boxed{x < 3}$

Step-by-step explanation:

4x - 1 < 11

Add 1 on both sides.

4x - 1 + 1 < 11 + 1

4x < 12

Divide both sides by 4.

(4x)/4 < 12/4

x < 3

8 0

3 years ago

Solve for x in the equation x squared + 2 x + 1 = 17.

Papessa [141]

Answer:

$x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\$

Step-by-step explanation:

given equation

$x^2 +2x +1 = 17$

subtracting 17 from both sides

$x^2 +2x +1 = 17\\x^2 +2x +1 -17= 17-17\\x^2 +2x - 16 = 0\\$

the solution for quadratic equation

$ax^2 + bx + c = 0$ is given by

x = $x = -b + \sqrt{b^2 - 4ac} /2a \\\\and \ \\-b - \sqrt{b^2 - 4ac} /2a$

________________________________

in our problem

a = 1

b = 2

c = -16

$x =( -2 + \sqrt{2^2 - 4*1*-16}) /2*1 \\x =( -2 + \sqrt{4 + 64}) /2\\x =( -2 + \sqrt{68} )/2\\x = ( -2 + \sqrt{4*17} )/2\\x = ( -2 + 2\sqrt{17} )/2\\x = - 1 + \sqrt{17}\\and\\\\x = - 1 - \sqrt{17}\\$

thus value of x is

$x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\$

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2 years ago