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madam [21]
3 years ago
12

HEEEELLLPP!!!

Mathematics
2 answers:
Arte-miy333 [17]3 years ago
6 0
The answer is 20 if the “18-2” is in the square root
If it’s square root of 18 and then minus 2
The answer is 15 square root 2 minus 2
Or 19.2132
Ray Of Light [21]3 years ago
3 0
The correct answer is 20 hope this helps
You might be interested in
Express 0.01620.01620, point, 0162 as a fraction.
just olya [345]
162/10000 is the answer to this question

8 0
3 years ago
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
Shkiper50 [21]

Answer:

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

Step-by-step explanation:

<u>Inequalities</u>

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

We must find the set of values for x that make the expression stand

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

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

\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 }

7 0
2 years ago
A teacher writes the following product on the board:
GREYUIT [131]

Answer:

Zhenghao is correct

Step-by-step explanation:

We have that:

(2x+3)(x + 4) = x² + 7x + 12

The only two factors of  x² + 7x + 12 different from 1 ,-1 and and itself are (2x+3) and (x + 4)

If Hafsana says that (x + 3) is a factor of x² + 7 + 12, He or she is not correct.

If Zhenghao says that x² + 7x + 12 is divisible by 2x + 3, He or she is correct because (2x + 3) is a factor of x² + 7x + 12

6 0
3 years ago
What is the total weight, w, of 5 bags of soil?
Charra [1.4K]

Answer:200

Step-by-step explanation: I typed this and got it right

4 0
2 years ago
Helpppppppppppppppppppppppppppppppppppppppppppp
Anestetic [448]

1.-36

2.-22.06

3.-42.1

You welcome :)


7 0
2 years ago
Read 2 more answers
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