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oksian1 [2.3K]
3 years ago
13

9log9(4) = A. 3 B. 4 C. 9 D. 81

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
5 0
"<span>9log9(4) = " What do you mean by this?</span>
You might be interested in
|
alekssr [168]

Answer: the answer is infinite solutions

Step-by-step explanation:Systems of Equations Level 1:

Question 1

Two linear equations are given below.

Exactly how many solutions does this system of equations have?

Select one:

4 0
3 years ago
What Is an exponent? ​
just olya [345]

Answer:

a quantity representing the power to which a given number or expression is to be raised, usually expressed as a raised symbol beside the number or expression (e.g. 3 in 23 = 2 × 2 × 2).

Step-by-step explanation:

6 0
3 years ago
There were 80 runners to start a race in the first half of the race, 1/5 of them dropped out. In the second half of the race, 1/
charle [14.2K]

Answer:

4

Step-by-step explanation:

8 0
3 years ago
Π π ππ π ππ π ππ π ππ π ππ π ππ π ππ π ππ π ππ π ππ π ππ π ππ π ππ π ππ π ππ π ππ π π
Serga [27]
The answer would be 1.405332e-13
Hope I helped!
7 0
3 years ago
Read 2 more answers
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
3 years ago
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