Ben bought 6 choclate csndies for $1.86.how many choclate candies can benn buy for 3.42

Mathematics

1 answer:

Irina-Kira [14]3 years ago

7 0

Answer: 7 chocolate candies

Step-by-step explanation: Ben would buy 7 chocolate candies for $3.42 because for 6 it would be $1.86 for 7 it would be $3.42.

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Evaluate the following

IRINA_888 [86]

(a) $[\frac{9}{2.6} - \frac{2.5^{2} }{2.5} ]^{2}$

Answer:

$[\frac{9}{2.6} - \frac{2.5^{2} }{2.5} ]^{2}$

= $[\frac{9}{2.6} - \frac{2.5*2.5 }{2.5} ]^{2}$

= $[\frac{9}{2.6} - \frac{2.5}{1} ]^{2}$

*canceling 2.5 in numerator and denominator*

$= [\frac{9-(2.5)(2.6)}{2.6} ]^2\\*Using L.C.M of 2.6 and 1 which comes out to be '2.6'= [\frac{9-(6.5)}{2.6} ]^2\\= [\frac{2.5}{2.6} ]^2\\*multiplying and dividing by '10'= [\frac{2.5*10}{2.6*10} ]^2\\= [\frac{25}{26} ]^2\\= \frac{25^2}{26^2}\\= \frac{625}{676}\\= 0.925$

Properties used:

Cancellation property of fractions

Least Common Multiplier(LCM)

The least or smallest common multiple of any two or more given natural numbers are termed as LCM. For example, LCM of 10, 15, and 20 is 60.

(b) $[[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3} ] ^{2}$

Answer:

$[[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3}] ^{2}\\$

*using $[x^{a}]^b = x^{ab}$ *

$= [\frac{3x^{3a}y^{3b}} {-3x^{3a} y^{3b} }] ^{2}$

*Again, using $[x^{a}]^b = x^{ab}$ *

$= \frac{3x^{2*3a}y^{2*3b}} {-3x^{2*3a} y^{2*3b} } \\= (-1)\frac{3x^{6a}y^{6b}} {3x^{6a} y^{6b} }\\[\tex]*taking -1 common, denominator and numerator are equal*[tex]= -(1)\frac{1}{1}\\= -1$