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3 years ago

Find the least common multiple (LCM) of 42, 50, and 54

Mathematics

2 answers:

Snowcat [4.5K]3 years ago

5 0

ANSWER:

LCM of 50, 42 and 54–

2 l 50 42 54

3 l 25 21 27

5 l 25 7 9

5 l 5 7 9

7 l 1 7 9

9 l 1 1 9

l 1 1 1

Multiply the divisors to find the LCM of 50, 42 and 54.

6 x 5 x 5 x 7 x 9 = 9450

LCM(30, 42, 54) = 9450.

Hence, The least common multiple for three numbers 30, 42 and 54 is 9450.

Anna007 [38]3 years ago

4 0

Answer:

LCM = 9450

Step-by-step explanation:

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2 years ago

What is the equation of a circle whose center at (-5, -2) and radius is 2?

aliya0001 [1]

The equation of a circle whose centre ( $-5,-2$ ) and radius $2$ is $x^{2}+y^{2}+10x+4y+25=0$

What is equation of a circle?

A circle is a closed curve drawn from a fixed point called centre of the circle. The distance between the centre of the circle and the arc of the circle is called the radius of the circle.

The equation of a circle with centre $(h, k)$ radius $r$ is

$(x-h)^{2} +(y-k)^{2}=r^{2}$

Given center = $(-5,-2)$ and Radius = $2$

Put the Given value in the equation:

= $(x-(-5))^{2} +(y-(-2))^{2} = 2^{2}$

= $(x+5)^{2} +(y+2)^{2}=2^{2}$

= $(x^{2} +10x+25)+(y^{2}+4y+4)= 4$

= $x^{2} +y^{2}+10x+ 4y+ 29 = 4$

= $x^{2} +y^{2}+10x +4y+ 29-4=0$

= $x^{2} +y^{2}+10x+4y+25=0$

So, the equation of the circle whose centre is $(-5,-2)$ and Radius $2$ is

$x^{2}+y^{2}+10x+4y+25=0$

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1 year ago

A plane flying a straight course observes a mountain at a bearing of 30.3degrees to the right of its course. At that time the pl

Svetllana [295]

Answer:

12km or 12 kilometers

Step-by-step explanation:

We are given the following values:

For the first bearing we have:

30.3 degrees at a distance of 9 kilometers

We are asked to find the second distance of the second bearing at 40.3 degrees

Therefore, we have:

30.3 degrees = 9km

40.3 degrees = ?? Unknown( we designate this as y)

We crossmultiply

30.3 degrees × y = 9km × 40.3 degrees

Divide the both sides by 30.3 degrees

y = (9km × 40.3 degrees) ÷ 30.3 degrees

y = 362.7/30.3 degrees

y = 11.97029703 km

Approximately to the nearest tenth of a km

y = 12km.

Therefore, the distance of the plane from the mountain when the second bearing is taken (to the nearest tenth of a km) is 12km.

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3 years ago