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mote1985 [20]
2 years ago
14

How to solve this? Is the question correct? ty :)

Mathematics
1 answer:
bezimeni [28]2 years ago
4 0

Answer:

The proposition is true.

Step-by-step explanation:

Now we proceed to demostrate that expression given is true by algebraic means:

1) \frac{x^{-1}+y^{-1}}{x^{-1}}+\frac{x^{-1}-y^{-1} }{y^{-1}} Given

2) \frac{y^{-1}\cdot (x^{-1}+y^{-1})+x^{-1}\cdot (x^{-1}-y^{-1})}{x^{-1}\cdot y^{-1}}   \frac{a}{b} + \frac{c}{d} = \frac{a\cdot d+b\cdot c}{b\cdot d}

3) \frac{x^{-1}\cdot y^{-1}+y^{-2}+x^{-2}-x^{-1}\cdot y^{-1}}{x^{-1}\cdot y^{-1}} Distributive property/a^{b}\cdot a^{c} = a^{b+c}

4) \frac{x^{-2}+y^{-2}}{(x\cdot y)^{-1}} Commutative, associative and modulative properties/Existence of additive inverse/a^{b}\cdot c^{b} = (a\cdot c)^{b}

5) [(x\cdot y)^{-1}]^{-1}\cdot (x^{-2}+y^{-2}) Commutative property/Definition of division

6) (x\cdot y)\cdot (x^{-2}+y^{-2})         (x^{-1})^{-1}

7) x^{-1}\cdot y + x\cdot y^{-1} Distributive property/Associative property/a^{b}\cdot a^{c} = a^{b+c}

8) (x^{-1}\cdot y^{-1})\cdot y^{2} + (x^{-1}\cdot y^{-1})\cdot x^{2}  Modulative property/Existence of additive inverse/a^{b}\cdot a^{c} = a^{b+c}

9) (x^{-1}\cdot y^{-1})\cdot (x^{2}+y^{2}) Distributive property

10) (x\cdot y)^{-1}\cdot (x^{2}+y^{2}) a^{b}\cdot c^{b} = (a\cdot c)^{b}

11) \frac{x^{2}+y^{2}}{x\cdot y} Commutative property/Definition of division/Result

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Consider the parent quadratic function f(x) = x2.
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Banabas must pay his ex-wife an amount of R350 000 in two years’ time. Calculate the amount that he must invest today to have th
White raven [17]

Answer:

He must invest R297 521 today.

Step-by-step explanation:

The compound interest formula is given by:

A(t) = P(1 + \frac{r}{n})^{nt}

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Banabas must pay his ex-wife an amount of R350 000 in two years’ time.

This means that t = 2, A(t) = 350000

Interest rate of 8.15% per annum compounded monthly:

This means that r = 0.0815, n = 12.

Amount he must invest today:

This is P. So

A(t) = P(1 + \frac{r}{n})^{nt}

350000= P(1 + \frac{0.0815}{12})^{2*12}

P = \frac{350000}{(1 + \frac{0.0815}{12})^{2*12}}

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He must invest R297 521 today.

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