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sukhopar [10]
3 years ago
12

Find the least common multiple of 8 and 6

Mathematics
2 answers:
Sedbober [7]3 years ago
8 0

To find the least common multiple of 6 and 8, first start by listing a few multiples of each number.

First, let's start by listing the multiples of 6.

1 × 6 = 6

2 × 6 = 12

3 × 6 = 18

4 × 6 = 24

Next, let's list the multiples of 8.

1 × 8 = 8

2 × 8 = 16

3 × 8 = 24

We can stop here because all other multiples that we find will be greater than 24. Therefore, the least common multiple of 6 and 8 is 24.

Zolol [24]3 years ago
4 0

the least common multiple of 8 and 6 is 24


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20% of a is 11. What is a?​
Reika [66]

Answer:

a=55

Step-by-step explanation:

This question is asking for you to find the whole (a), given a part (11). This can be done through multiplication. 20% is 1/5 of 100%. This means to find the whole number, you should multiply the part by 5. 11*5=55. So, a must equal 55.

We can check this answer by working backward. Do this by finding 20% of 55, and checking to see if it is 11. To find a percentage of a number, multiply that number by the percent in decimal form. So, 20% of 55 is equal to 0.2*55, which is 11. Therefore, the answer is correct.

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3 years ago
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دائرة مركزها o فيها الوتران cd,ab اذا علمت ان النقطة n نقطة منتصف الوتر cd والنقطة m نقطة منتصف الوترab وكان om=on وطول cn=6cm
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Yuma needs a singer. Singer A is offering her services for an initial $75 in addition to $35 per hour. Singer B is offering his
Bingel [31]

Answer:

5 hours

Step-by-step explantioon:

35x+75=20x+150

15x+75=150

15x=75

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6 0
3 years ago
Find four distinct complex numbers (which are neither purely imaginary nor purely real) such that each has an absolute value of
Luda [366]

Answer:

  • 0.5 + 2.985i
  • 1 + 2.828i
  • 1.5 + 2.598i
  • 2 + 2.236i

Explanation:

Complex numbers have the general form a + bi, where a is the real part and b is the imaginary part.

Since, the numbers are neither purely imaginary nor purely real a ≠ 0 and b ≠ 0.

The absolute value of a complex number is its distance to the origin (0,0), so you use Pythagorean theorem to calculate the absolute value. Calling it |C|, that is:

  • |C| = \sqrt{a^2+b^2}

Then, the work consists in finding pairs (a,b) for which:

  • \sqrt{a^2+b^2}=3

You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:

\sqrt{a^2+b^2}=3\\ \\ a^2+b^2=3^2\\ \\ a^2=9-b^2\\ \\ a=\sqrt{9-b^2}

I will use b =0.5, b = 1, b = 1.5, b = 2

b=0.5;a=\sqrt{9-0.5^2}=2.958\\ \\b=1;a=\sqrt{9-1^2}=2.828\\ \\b=1.5;a=\sqrt{9-1.5^2}=2.598\\ \\b=2;a=\sqrt{9-2^2}=2.236

Then, four distinct complex numbers that have an absolute value of 3 are:

  • 0.5 + 2.985i
  • 1 + 2.828i
  • 1.5 + 2.598i
  • 2 + 2.236i
4 0
3 years ago
.If the side of an equilateral triangle is 6cm, then the area of triangle is
Oksi-84 [34.3K]

Answer:

9\sqrt{3} cm^2

Step-by-step explanation:

area of an equilateral triangle = \frac{s^{2} \sqrt{3} }{4} (s = length of the side)

area = \frac{6^{2}\sqrt{3}  }{4}

area = \frac{36\sqrt{3} }{4}

area = 9\sqrt{3} cm^2

4 0
3 years ago
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