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3 years ago

Topic: Simultaneous equations Class 9 ICSE

Mathematics

1 answer:

BARSIC [14]3 years ago

8 0

Denote the number by 10a + b, where both a and b are integers picked from {1, 2, 3, …, 9}.

"the difference of whose digits is 3" ==> |a - b| = 3

"4 times the number is equal to 7 times the number obtained by reversing the digits" ==> 4 (10a + b) = 7 (10b + a)

Simplifying the second equation, you get

40a + 4b = 70b + 7a

33a - 66b = 0

a - 2b = 0

Assume a > b, in which case we would have |a - b| = a - b, so

a - b = 3

Eliminate a to solve for b, then for a :

(a - b) - (a - 2b) = 3 - 0

b = 3 ==> a = 6

Then the original number is 63.

If instead we had assumed a < b, we would have had |a - b| = b - a = 3. Then

2 (b - a) + (a - 2b) = 2×3 + 0

-a = 6

But neither a nor b can be negative, so this case is moot.

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3 years ago

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Andrews [41]

Answer:

$f(x) = \frac{24}{25} * \frac{5}{6}^x$

Step-by-step explanation:

Given

$(x_1,y_1) = (2,\frac{2}{3})$

$(x_2,y_2) = (3,\frac{5}{9})$

Required

Write the equation of the function $f(x) = ab^x$

Express the function as:

$y = ab^x$

In: $(x_1,y_1) = (2,\frac{2}{3})$

$y = ab^x$

$\frac{2}{3} = a * b^2$ --- (1)

In $(x_2,y_2) = (3,\frac{5}{9})$

$y = ab^x$

$\frac{5}{9} = a * b^3$ --- (2)

Divide (2) by (1)

$\frac{5}{9}/\frac{2}{3} = \frac{a*b^3}{a*b^2}$

$\frac{5}{9}/\frac{2}{3} = b$

$\frac{5}{9}*\frac{3}{2} = b$

$\frac{5}{3}*\frac{1}{2} = b$

$\frac{5}{6} = b$

$b = \frac{5}{6}$

Substitute 5/6 for b in (1)

$\frac{2}{3} = a * b^2$

$\frac{2}{3} = a * \frac{5}{6}^2$

$\frac{2}{3} = a * \frac{25}{36}$

$a = \frac{2}{3} * \frac{36}{25}$

$a = \frac{2}{1} * \frac{12}{25}$

$a = \frac{24}{25}$

The function: $f(x) = ab^x$

$f(x) = \frac{24}{25} * \frac{5}{6}^x$

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3 years ago

Topic: Simultaneous equations Class 9 ICSE​

Topic: Simultaneous equations Class 9 ICSE