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iren2701 [21]
3 years ago
15

Working together, Katherine and Marina can sand a large cabinet in 2 hours. It would take Katherine 10 hours to do the job alone

. What is the value of r, the part of the job that Marina can complete in 1 hour?
Mathematics
2 answers:
Sauron [17]3 years ago
3 0

The <em>correct answer</em> is:

0.4 of the job.

Explanation:

Since Katherine can complete the job (100%, or 1) in 10 hours, she can complete 1/10 of it in 1 hour.

Both women together can complete the job in 2 hours. This means Katherine's rate multiplied by 2 hours, 0.1(2), added to Marina's rate multiplied by 2 hours, r*2, equal the entire job (100%, or 1 whole):

0.1(2)+r(2) = 1

0.2+2r = 1

Subtract 0.2 from each side:

0.2+2r-0.2 = 1-0.2

2r = 0.8

Divide each side by 2:

2r/2 = 0.8/2

r = 0.4

Marina's rate is 0.4 of the job per hour.

lakkis [162]3 years ago
3 0
So, Katherine can do 0.1 of the job per hour. (if she needs 10 hours for the job, each hour she can do the whole divided by the time, so 1/10, which is 0.1)

if in two hours she and Marina can be done, then Marina will do 1(the whole job)-2/10 (twice as much as she can do in an hour)=0.8 of the job. So if Marina does 0.8 of the job in two hours, each hour she will do half of it: 0.4 - and this is the correct answer

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3 years ago
Give an example of a function f : N → N that is surjective but not injective. You must explain why your example is surjective an
Helen [10]

Answer:

Consider f: N → N defined by f(0)=0 and f(n)=n-1 for all n>0.

Step-by-step explanation:

First we will prove that f is surjective. Let y∈N be any natural number. Define x as the number x=y+1. Then x∈N, and f(x)=x-1=(y+1)-1=y.  We conclude that f is surjective.

However, f is not injective. Take x1=0 and x2=1. Then x1≠x2 but f(x1)=0 and f(x2)=x2-1=1-1=0. We have shown that there are two natural numbers x1,x2 such that x1≠x2 but f(x1)=f(x2), that is, f is not injective.

Note:

If 0∉N in your definition of natural numbers, the same reasoning works with the function f: N → N defined by f(1)=1 and f(n)=n-1 for all n>1. The only difference is that you consider x1=1, x2=2 for the injectivity.

7 0
3 years ago
The region bounded by y=(3x)^(1/2), y=3x-6, y=0
Ganezh [65]

Answer:

4.5 sq. units.

Step-by-step explanation:

The given curve is y = (3x)^{\frac{1}{2} }

⇒ y^{2} = 3x ...... (1)

This curve passes through (0,0) point.

Now, the straight line is y = 3x - 6 ....... (2)

Now, solving (1) and (2) we get,

y^{2} - y - 6 = 0

⇒ (y - 3)(y + 2) = 0

⇒ y = 3 or y = -2

We will consider y = 3.

Now, y = 3x - 6 has zero at x = 2.

Therefor, the required are = \int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx

= \sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}

= [\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} }  }{3}] - [13.5 - 18 - 6 + 12]

= 6 - 1.5

= 4.5 sq. units. (Answer)

7 0
3 years ago
Please help me with question 24 ‍♀️
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3 0
3 years ago
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answer is 16
4 0
3 years ago
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