What is the least common muliple of 5 and 6

2 answers:

8 0

Answer: The LCM of 15 and 6, so the least common multiple, the smallest multiple that they have in common, we see over here. 15 times 2 is 30. And 6 times 5 is 30.

lutik1710 [3]3 years ago

8 0

The least common multiple would be 30

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There were 230,600 jobs available in the field of radiology in the year 2014. Each year, that number is expected

vladimir1956 [14]

Answer:

J(t) = 230,600(1.009)^t

Step-by-step explanation:

J(t) = 230,600(1 + 0.009)^t, or

J(t) = 230,600(1.009)^t

6 0

3 years ago

___ x 4 = 12<br> ___ x 5 = 20<br> ___ x 2 = 10

RideAnS [48]

Answer:

3 x 4 = 12

4 x 5 = 20

5 x 2 = 10

Step-by-step explanation:

3 + 3 + 3 + 3 = 12

3 four times equals 12

This is the same for the others

4 + 4 + 4 + 4 + 4 = 20

Four 5 times equals 20

4 0

3 years ago

I will mark brainliest IF CORRECT

umka2103 [35]

Answer:

1 - (π/6) - (√3/4)

Step-by-step explanation:

First, calculate the area of ABCD. That's just $1*1 = 1 un^{2}$

Then, we are told that arcs BD and AC are circular, meaning they are portions of a circle. Therefore, let's redraw this image a little. Look below for the image.

Essentially, what I've done is created an equilateral triangle and two sectors. You may be wondering why the triangle is equilateral. Well, we firstly know it's isosceles because the two original arcs are circular and congruent. Their overlap creates two congruent lines that can be extend from point M to the vertices A and D. These two lines also happen to be radii of the quarter circles because the two sectors have AM AND AD as radii. AM and AD = 1, so AMD is an equilateral triangle.

Draw altitude OD. This is a right angle. Since equilateral triangles are also isosceles triangles, and isosceles triangles have the property of sharing the median to base and altitude to base, we can say that OD = $\frac{1}{2}$ AD = $\frac{1}{2}$ . MD = 1, so this triangle is a 30-60-90 triangle. Why? This is because of the 30º angle converse theorem. The shorter leg is half of the hypotenuse MD, so <OMD is a 30º angle, making <ODM a 60º angle and inevitably <MDC a 30º angle (because <D is a right angle) and <ABM as well because the two sectors are congruent.

Because m<ABM = m<MDC = 30º, these two sectors are 30/360 of the individual circles or 1/12 of the original circles, which have areas of π (because 1^2*π = π). Each is 1/12π, combined they are 1/6π.

The area of the equilateral triangle is √3/8. This is because MO can be calculated, using the 30-60-90 ratios, to be √3/2. (1 * √3/2)/2 = (√3/2)/2 = √3/4; this uses the area of a triangle. Now, time to calculate the shaded area.

The area of the square is 1, the area of the sectors π/6, and the equilateral triangle √3/4. Subtract the equilateral triangle and sectors' areas from the area of the square to be left with the remaining shaded part. 1-(π/6)-(√3/4) is your final answer. To put that a little more cleanly below in the image.

Hope this helps, have a great day.