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3 years ago

What table of values goes with the equation y = l-2x l?

Mathematics

1 answer:

Sergio [31]3 years ago

6 0

x, y

0, 0

1, 2

2, 4

3, 6

4, 8

etc.

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In 16 years, Ben will be 3 times as old as he is right now.How old is he right now?

Goryan [66]

Ben is 8 years old right now.

7 0

3 years ago

Read 2 more answers

Let omega be a complex number such that omega^3 = 1. Find all possible values of {1}{1 + \omega} + {1}{1 + \omega^2}. Enter all

Norma-Jean [14]

Answer:

1 is Answer.

Step-by-step explanation

$\frac{1}{1+omega^{2} } + \frac{1}{1+omega }\\$

= $\frac{(-1)*(omega + 1)}{omega^{2} }$

As we know that ω²+ω+1=0

Thus putting in above equation, we get

= $\frac{1}{(-1)*omega } + \frac{1}{(-1)*omega^{2} }$

Rearranging and simplifying:

= $\frac{-1}{omega } + \frac{-1}{omega^{2} }$

= $\frac{(-1)*(omega + 1)}{omega^{2} }$

= $\frac{(-1)*(- omega^{2} )}{omega^{2} }$

= 1 Answer

8 0

3 years ago

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

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