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vovikov84 [41]
3 years ago
13

⦁ If we add 5 to the denominator and subtract 5 from the numerator of a fraction, it becomes . If we subtract 3 from the numerat

or and add 3 to its denominator, it reduces to . Find the fraction.correct will win the brainliest
Mathematics
1 answer:
natima [27]3 years ago
5 0

the complete question is

If we add 5 to the denominator And subtract 5 from the numerator of a fraction ot reduces to 1/7. if we subtract 3 from the numerator and add 3 to its denominator ot reduces to 1/3. find the fraction.


Let

x/y-----------> the fraction


we know that

(x-5)/(y+5)=1/7------>7*(x-5)=(y+5)-----> y=7x-35-5----> y=7x-40----> equation 1

(x-3)/(y+3)=1/3----> 3*(x-3)=y+3----> y=3x-9-3----> y=3x-12------> equation 2


equate equation 1 and equation 2

7x-40=3x-12-----> 7x-3x=40-12------> 4x=28-----> x=7


y=3x-12------> y=3*7-12-----> y=9

x/y=7/9


the answer is

the fraction is 7/9

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The ratio of boys to girl in a school is 5:7 if there are 600 students, how many girls are there
larisa86 [58]

Answer:

350 girls

Step-by-step explanation:

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The secret to solving problems with ratios is to find the value of one unit.

5:7 = 12 units total

To find one unit, divide the total number of students by the total number of units.

600/12 = a

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3 0
2 years ago
If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A​
anygoal [31]

Answer:

The value of the given expression is

\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

Step by step Explanation:

Given that tanA=a

To find the value of \frac{sin4A-2sin2A}{sin4A+2sin2A}

Let us find the value of the expression :

\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A} ( by using the formula sin2A=2cosAsinA here A=2A)  

=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}

=\frac{(cos2A-1)}{(cos2A+1)}

=\frac{(-(1-cos2A))}{(1+cos2A)}(using  sin^2A+cos^2A=1  here A=2A)

=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}(using cos2A=cos^2A-sin^2A here A=2A)

=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}

=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}

=\frac{-2sin^2A}{2cos^2A}

=-\frac{sin^2A}{cos^2A}

=-tan^2A  ( using tanA=\frac{sinA}{cosA} here A=2A )

 =-a^2 (since tanA=a given )

Therefore \frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

6 0
3 years ago
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