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3 years ago

Elam can wire a light in 40

Mathematics

2 answers:

NeX [460]3 years ago

6 0

Answer:

24 minutes

Step-by-step explanation:

We use 1/a + 1/b = 1/c

where a and b are the individual times and c is the time together

1/40 + 1/60 = 1/c

Multiply each side by 120c

120c( 1/40 + 1/60 = 1/c)

3c + 2c = 120

5c = 120

Divide each side by 5

5c/5 = 120/5

c = 24

zvonat [6]3 years ago

5 0

Answer: 24 minutes

Step-by-step explanation: To solve this kind of a problem which is called a work problem, it's important to understand the following idea.

Since Elam can wire a light in 40 minutes, we know that in 1 minute,

Elam can wire 1/40 of the light and in 2 minutes, Elam can wire 2/40 of it.

Therefore, in t hours, Elam can wire t/40 of the light.

The same applies for Michael, in t hours, he can wire t/60 of the light.

So to solve the problem, we use the following formula:

Part of job done by Elam + part of job done Michael = 1 job done

The part of the job done by Elam is t/40 and t/60 for Michael.

So we have t/40 + t/60 = 1.

Now multiply both sides by 120 to get 3t + 2t = 120

or 5t = 120 so t = 24.

So it takes 24 minutes if they work together.

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4 years ago

Read 2 more answers

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi

natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

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This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$ ,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

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Step-by-step explanation:

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