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Musya8 [376]
3 years ago
13

3(x + 1) = 5(x - 2) + 7 x=?

Mathematics
2 answers:
Rom4ik [11]3 years ago
6 0
Those are called binomials & you have to use the distributive property.
first: 3(x+1) distributed & simplified 3x+3 & 5(x-2) distributed & simplified is 5x-10, therefore 3(x+1) = 3x+3 & 5(x-2) = 5x-10. always simplify before solving any algebraic problem! 
second: so now we can say... 3x + 3 = 5x - 10 + 7 .
third: now we combine like terms by adding the -10 & 7 (on the right side) together to get -3.
fourth: now we have to isolate the variables by doing the opposite function on each side! it should look like this right now: 3x + 3 = 5x - 3 ... subtract the 3x, on both sides, from the 5x to get 2x. looks like this now: 3 = 2x - 3
now add the -3 (on the right) on the both sides
looks like this now: 6 = 2x
fifth: now divide to get 3.     X = 3 !!!!!

alukav5142 [94]3 years ago
4 0
1.4 repeating? 


3x+3=5x-10+7x
3x+3=12x-10
3=9x-10
13=9x
1.4 repeating=x
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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}\\----------------------

Hope~This~Helped!~Good~Luck!

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