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Mrac [35]
3 years ago
14

Plz help..................................... ​

Mathematics
1 answer:
Andre45 [30]3 years ago
7 0

7

7x5=35

7x5^2=175

5x5=25

7x5^3=875

5x5x5=125

7x5^4=4375

5x5x5x5=625

add them :>

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What is the prime factorization for 36?<br> A. 22x32<br> B. 2x3x6<br> C. 6x6
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Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
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\displaystyle  -\frac{1}{2} \leq x < 1

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<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

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\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

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\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0

Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

(6x^3+14x^2+10x+12)

is always positive and doesn't affect the result. It can be neglected. The expression

(x-\frac{1}{2})^2

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0

For the expression to be negative, both signs must be opposite, that is

(x+\frac{1}{2})\geq 0, (x-1)

Or

(x+\frac{1}{2})\leq 0, (x-1)>0

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

\displaystyle  -\frac{1}{2} \leq x < 1

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

\boxed{\displaystyle  -\frac{1}{2} \leq x < 1 }

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