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Mumz [18]
3 years ago
15

Three workers have to do a certain job. The first worker can finish the job in 8 hours. The second worker can finish the job in

4 hours. The third worker can also finish the job in 4 hours. How long will it take the three workers to finish the job if they work together?
PLS help
Mathematics
1 answer:
mestny [16]3 years ago
8 0

Working together three workers would take 1 hour 36 minutes to finish the job

<em><u>Solution:</u></em>

Given that first worker can finish the job in 8 hours

So in one hour, first worker can do \frac{1}{8} th of the work

The second worker can finish the job in 4 hours

So in one hour, second worker can do \frac{1}{4} th of the work

The third worker can also finish the job in 4 hours

So in one hour, third worker can do \frac{1}{4} th of the work

<em><u>The three workers working together in 1 hour can do:</u></em>

\frac{1}{8} + \frac{1}{4} + \frac{1}{4} = \frac{1 + 2 + 2}{8} = \frac{5}{8}

The three worker can thus do \frac{5}{8} th of the work in one hour

Hence the three of them together can finish the work in \frac{8}{5} hours

\frac{8}{5} = 1\frac{3}{5} hours

Thus working together three workers would take 1 hour 36 minutes to finish the job

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5 0
3 years ago
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The speed of an object in space is shown in the graph.
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Step-by-step explanation:

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so, we have 2 points here

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x (or time) changes by 0.2 units (from 0.2 to 0.4).

y (or distance) changes by 3 units (from 3 to 6).

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Solve x2 − 8x + 15 &lt; 0.
n200080 [17]

Answer:

Step-by-step explanation:

Given the quadratic inequality we are to find the interval of x

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x²-3x-5x+15<0

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3 years ago
1 (a) (i) Yasmin and Zak share an amount of money in the ratio 21 : 19.
cupoosta [38]

Answer: 1(a) $120

(ii a) $34

Step-by-step explanation:

21-19 = 2

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21x 3( one ratio =$3) =63

19x3 = 57

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3 years ago
Find the perimeter of a field that has length 2/x + 1 and width 5/x^2 -1.
Semmy [17]

Answer:

D)4x + 6/(x + 1)(x - 1)

Step-by-step explanation:

A field is basically a rectangle, so to find the perimeter of our field we are using the formula for the perimeter of a rectangle

p=2(l+w)

where

p is the perimeter

l is the length

w is the width

We know from our problem that the field has length 2/x + 1 and width 5/x^2 -1, so l=\frac{2}{x+1} and w=\frac{5}{x^2-1}.

Replacing values:

p=2(l+w)

p=2(\frac{2}{x+1} +\frac{5}{x^2-1})

Notice that the denominator of the second fraction is a difference of squares, so we can factor it using the formula a^2-b^2=(a+b)(a-b) where a is the first term and b is the second term. We can infer that a=x^2 and b=1^2. So, x^2-1=(x+1)(x-1). Replacing that:

p=2(\frac{2}{x+1} +\frac{5}{x^2-1})

p=2(\frac{2}{x+1} +\frac{5}{(x+1)(x-1})

We can see that the common denominator of our fractions is (x+1)(x-1). Now we can simplify our fraction using the common denominator:

p=2(\frac{2(x-1)+5}{(x+1)(x-1)} )

p=2(\frac{2x-2+5}{(x+1)(x-1)} )

p=2(\frac{2x+3}{(x+1)(x-1)} )

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We can conclude that the perimeter of the field is D)4x + 6/(x + 1)(x - 1).

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4 years ago
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