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3 years ago

Which is bigger 2/3 or 1/3?

Mathematics

2 answers:

Lyrx [107]3 years ago

5 0

2/3 is bigger than 1/3......TO CHECK this answer 2/3=66% or .66 and 1/3=33% or .33

Therefore, 2/3(66%) is greater than 1/3(33%)

2/3>1/3

Paha777 [63]3 years ago

4 0

2/3 Is bigger
2/3 = .66
1/3 = .33
66 is larger than 33.

Fractons that are bigger than 1/3 are: 2/3 3/3 4/3

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$Hey~freckledspots!\\----------------------$

$We~will~solve~for~i^{425}!$

$Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\$

$Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------$

$Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------$

$Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------$

$Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------$

$Answer:\\\large\boxed{-1+i}\\----------------------$

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3 years ago

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Let f (x ) = 3x² - 4 and let g (x) = 2x + 1. <br> Find the value of f[g(2)].

stepan [7]

Answer:

The value of f[g(2)] is 71.

Step-by-step explanation:

We need to find f[g(2)].

The first step is finding f[g(x)]

We have that:

$f(x) = 3x^{2} - 4$

$g(x) = 2x + 1$

To find f[g(x)], the first step is replacing each x is f by g(x). So

$f(g(x)) = 3(2x+1)^{2} - 4$

Now we replace x by 2

$f(g(2)) = 3(2*2+1)^{2} - 4 = 75 - 4 = 71$

The value of f[g(2)] is 71.

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