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Phoenix [80]
3 years ago
9

In a lighting, green light glows after every 12 seconds, red light glows after every 30 seconds, blue light glows after every 34

seconds and yellow light glows after every 56 seconds. Find the least time in minutes after which they all glow together.​
Mathematics
1 answer:
AlexFokin [52]3 years ago
5 0

Answer:

238 minutes

Step-by-step explanation:

We are told:

In a lighting, green light glows after every 12 seconds, red light glows after every 30 seconds, blue light glows after every 34 seconds and yellow light glows after every 56 seconds.

We solve this question by using LCM Method

We find the Least Common Multiple of 12, 30, 36 and 56

= 14,280 seconds.

Hence, the least time in seconds after which they glow together is 14,280 seconds.

To find the least time in minutes after which they all glow together.​

We convert from seconds to minutes

60 seconds = 1 minute

14280 seconds = x

Cross Multiply

60 × x = 14280 × 1

x = 14280/60

x = 238 minutes

The least time in minutes after which they all glow together.​

238 minutes

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4/5mm what are the perimeter and the area of the square<br>​
Troyanec [42]

length+ breadth and 2×side

Step-by-step explanation:

4÷2 =2

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5 0
3 years ago
MZJ and m M are base angles of isosceles trapezoid JKLM. If mZJ = 18x + 8 and m_M=11x + 15 find m2K.
maxonik [38]

The question is incomplete. Here is the complete question.

m∠J and m∠Kare base angles of an isosceles trapezoid JKLM.

If m∠J = 18x + 8, and m∠M = 11x + 15 , find m∠K.

A. 1

B. 154

C. 77

D. 26

Answer: B. m∠K = 154

Step-by-step explanation: <u>Isosceles</u> <u>trapezoid</u> is a parallelogram with two parallel sides, called Base, and two non-parallel sides that have the same measure.

Related to internal angles, angles of the base are equal and opposite angles are supplementary.

In trapezoid JKLM, m∠J and m∠M are base angles, so they are equal:

18x + 8 = 11x + 15

7x = 7

x = 1

Now, m∠K is opposite so, they are supplementary, which means their sum results in 180°:

m∠J = 18(1) + 8

m∠J = 26

m∠K + m∠J = 180

m∠K + 26 = 180

m∠K = 154

The angle m∠K is 154°

7 0
3 years ago
points C,D, and E are collinear on CE, and CD:DE = 3/5. C is located at (1,8), D is located at (4,5), and E is located at (x,y).
kap26 [50]

Answer:

The point E is located at (9,0)

x=9, y=0

Step-by-step explanation:

we have that

Points C,D, and E are collinear on CE

Point D is between point C and point E

we know that

CE=CD+DE -----> equation A (by addition segment postulate)

\frac{CD}{DE}=\frac{3}{5}

CD=\frac{3}{5}DE ------> equation B

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

<em>Find the distance CD   </em>

we have

C(1,8), D(4,5)

substitute in the formula

CD=\sqrt{(5-8)^{2}+(4-1)^{2}}

CD=\sqrt{(-3)^{2}+(3)^{2}}

CD=\sqrt{18}\ units

<em>Find the distance DE</em>

substitute the value of CD in the equation B and solve for DE

\sqrt{18}=\frac{3}{5}DE

DE=\frac{5\sqrt{18}}{3}\ units

<em>Find the distance CE</em>

CE=CD+DE

we have

DE=\frac{5\sqrt{18}}{3}\ units

CD=\sqrt{18}\ units

substitute the values in the equation A

CE=\sqrt{18}+\frac{5\sqrt{18}}{3}

CE=\frac{8\sqrt{18}}{3}

<em>Applying the formula of distance CE</em>

we have

CE=\frac{8\sqrt{18}}{3}

C(1,8), E(x,y)    

substitute in the formula of distance

\frac{8\sqrt{18}}{3}=\sqrt{(y-8)^{2}+(x-1)^{2}}

squared both sides

128=(y-8)^{2}+(x-1)^{2}  -----> equation C

<em>Applying the formula of distance DE</em>

we have

DE=\frac{5\sqrt{18}}{3}\ units

D(4,5), E(x,y)    

substitute in the formula of distance

\frac{5\sqrt{18}}{3}=\sqrt{(y-5)^{2}+(x-4)^{2}}

squared both sides

50=(y-5)^{2}+(x-4)^{2}  -----> equation D

we have the system

128=(y-8)^{2}+(x-1)^{2}  -----> equation C

50=(y-5)^{2}+(x-4)^{2}  -----> equation D

Solve the system by graphing

The intersection point both graphs is the solution of the system

The solution is the point (9,0)

therefore

The point E is located at (9,0)

see the attached figure to better understand the problem

6 0
3 years ago
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