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3 years ago

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

2 answers:

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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alukav5142 [94]

Answer: The answer is 12

Step-by-step explanation: because 5+5+2=12.

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3 years ago

Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify yo

Rama09 [41]

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

$u\neq 0_{R2}$

$v\neq 0_{R2}$

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to $R^{2x2}$ using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

$A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]$

We used the first vector ''u'' as the first column of the matrix A

We used the second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

$Det(A)=6-p$

We need this determinant to be different to zero

$6-p\neq 0$

$p\neq 6$

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that $p\neq 6$

We can write : p ∈ IR - {6}

Notice that is $p=6$ ⇒

$u=(3,6)$

$v=(1,2)$

If we write $3v=3(1,2)=(3,6)=u$ , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.