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EastWind [94]
3 years ago
11

If the product of two positive fractions a and b is 15/56, find three pairs for possible values for a and b

Mathematics
2 answers:
Elina [12.6K]3 years ago
5 0
All you have to do is find pairs of factor for both 15 and 56
e.g. 1,15   3,5 
      1.56    2, 28   4, 14

so three possibilities could be
1/1 x 15/56
3/4 x 5/14 
1/2 x 15/28
love history [14]3 years ago
5 0

Answer

Find out the three pairs for possible values for a and b .

To prove

As given

The\ product\ of\ two\ positive\ fractions\ a\ and\ b\ is\ \frac{15}{56} .

i.e it is written in the form.

a\times b = \frac{15}{56}

When

a = \frac{3}{2} , b = \frac{5}{28}

Than

\frac{3\times 5}{2\times 28} = \frac{15}{56}

When

a = \frac{3}{4} , b = \frac{5}{14}

Than

\frac{3\times 5}{4\times 14} = \frac{15}{56}

When

a = \frac{1}{2} , b = \frac{15}{28}

Than

\frac{1\times 15}{2\times 28} = \frac{15}{56}

Hence proved



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Question 1 (Essay Worth 10 points) (01.02 MC) Part A: If (26)x = 1, what is the value of x? Explain your answer. (5 points) Part
katen-ka-za [31]

Answer:

See Explanation

Step-by-step explanation:

The question is not clear. However, I will treat the question as:

(26)x = 1

(50)x = 1

and:

(2^6)^x = 1

(5^0)^x = 1

Solving: (26)x = 1  and  (50)x = 1

(26)x = 1

Divide both sides by 26

x = \frac{1}{26}

(50)x = 1

Divide both sides by 50

x = \frac{1}{50}

Solving (2^6)^x = 1 and (5^0)^x = 1

(2^6)^x = 1

Express 1 as 2^0

(2^6)^x = 2^0

Remove bracket

2^{6x} = 2^0

Cancel out 2

6x = 0

Divide both sides by 6

x = \frac{0}{6}

x = 0

(5^0)^x = 1

Express 1 as 5^0

(5^0)^x = 5^0

Cancel out 5^0

x = 1

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3 years ago
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