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3 years ago

Find the least common denominator (LCD) of 1/20 and 11/30

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

5 0

Answer:

Step-by-step explanation:

To find the LCD, all you have to do is find the least common multiple (LCM) of the two denominators.

The fractions are already reduced, so you have to find the LCM of 20 and 30. The LCM of 20 and 30 is 60.

To find the LCM:

20 and 30 are easy numbers, so you can just list the multiples out.

20: 20, 40, 60, etc.

30: 30, 60

The lists both have 60, so the LCM is 60.

Another way is to prime factorize the two numbers.

20: 2^2*5

30: 2*3*5

Then, take the non-repeating factor from the second number (3) and multiply that by 30, which is 60.

Mrrafil [7]3 years ago

3 0

Answer:

Step-by-step explanation:

LCD(1 /20 , 11 /30) = LCM(20, 30) = 22×3×5 = 60

1 /20 = 3 /60

11 /30 = 22 /60

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36m⁴-√16 pls help asap

shusha [124]

Answer:

Step-by-step explanation:

$(a^2-b^2)=(a-b)(a+b)\\ \\ 36m^4-4\\ \\ (6m^2-2)(6m^2+2)\\ \\ 4(3m^2-1)(3m^2+1)$

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3 years ago

8.3 is equal to tenths hundredths

irakobra [83]

Answer:

83 Tenths

830 hundreths

Step-by-step explanation:

8.30=8.3 Remove the points and they will give you how many tenths or hundreths

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6 0

3 years ago

Read 2 more answers

If the area of AABC is D, give the expressions that complete the equation for the measure of ZB?

Ira Lisetskai [31]

Given:

The figure of triangle ABC.

The area of the triangle ABC is D.

$m\angle B=\sin ^{-1}(\dfrac{m}{n})$

To find:

The value of m and n in the given expression.

Solution:

Let h be the height of the triangle ABC.

Area of a triangle is:

$Area=\dfrac{1}{2}\times base\times h$

Where, b is the base and h is the height of the triangle.

$Area=\dfrac{1}{2}\times a\times h$

The area of the triangle ABC is D.

$D=\dfrac{1}{2}\times a\times h$

$2D=ah$

$\dfrac{2D}{a}=h$ ...(i)

In a right angle triangle,

$\sin \theta =\dfrac{Perpendicular}{Hypotenuse}$

$\sin B =\dfrac{h}{c}$

$\sin B =\dfrac{1}{c}\times \dfrac{2D}{a}$ [Using (i)]

$\sin B =\dfrac{2D}{ac}$

$m\angle B =\sin ^{-1}\dfrac{2D}{ac}$ ...(ii)

We have,

$m\angle B=\sin ^{-1}(\dfrac{m}{n})$ ...(iii)

On comparing (ii) and (iii), we get

$m=2D$

$n=ac$

Therefore, the required values are $m=2D, n=ac$ .

6 0

3 years ago

The following blank indicates a missing operation in the equation.

Natalija [7]

The answer would be 5/2q+9r+6r-20s

Explanation
We have the equation
5(1/2q+3r-4s)
which would equal
5/2q+15r-20s

so with the information
We can subtract 6r from 15r to get 9r
15r-6r=9r
+9r would be the missing piece

So we substitute it in to make the equation true.
5/2q(+9r)+6r-20s
Hope this helps or was correct

3 0

3 years ago

NEED HELP The value of x is .

Effectus [21]

The sum of all exterior angles of ANY polygon is equal to 360 degrees.

SO in this figure...
360 = 134 + 130 + x
360 = 264 + x
x = 360 - 264 = 96 degrees

3 0

3 years ago