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lions [1.4K]
3 years ago
8

Factor the polynomial completely. x2 - 3x + 1 A) (x + 1)(x + 1) B) (x - 1)(x - 1) C) (x - 3)(x + 1) D) cannot be factored

Mathematics
1 answer:
omeli [17]3 years ago
6 0
<span>D) cannot be factored
.............................................</span>
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The fish population in a pond evolves according to the logistic equation and a percentage of fish are removed for each unitof ti
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Step-by-step explanation:

Suppose that the carrying capacity is 900 fish, the intrinsic growth rate is equal to 2, and the harvest rate is 20%.

a. Use the graphical or eigenvalue approach

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Round 24,347 to the nearest thousand.
hodyreva [135]

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One base of a trapezoid is five times as long as the other. The height is the average of the two bases. The area is 441 square u
Gelneren [198K]

Answer:

The length of the longer base he 35 units

Step-by-step explanation:

Here, we want to find the length of the longer base of the trapezoid

Mathematically, we can find the area using the formula;

1/2( a + b) h

where a is the shorter base

b is the longer base

h is the height

Let the shorter base be x

The other base is 5 times this length and that makes 5 * x = 5x

Height is the average of both bases;

(x + 5x)/2 = 6x/2 = 3x

Substituting these in the formula, we have ;

1/2(x + 5x)3x = 441

3x(6x) = 882

18x^2 = 882

x^2 = 882/18

x^2 = 49

x^2 = 7^2

x = 7

But the longer base is 5x and that will be 5 * 7 = 35 units

4 0
3 years ago
PLEASE HELP I'LL MARK BRAINLIEST!!!!!
WARRIOR [948]
The answer is B And C
7 0
3 years ago
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Which is the graph of the function f(x)= negative square root x
Vsevolod [243]
First, note that \sqrt{x} is always positive (except for x=0), so -\sqrt{x} must be always negative.

Thus, the only plausible graphs are 1 and 3 since they are below the x-axis.

Now, \sqrt{x} and -\sqrt{x} are only defined for x≥0, because only for these x'es we can take the square root. 

Note that the third graph has domain (-infinity, 0], so it is not the right one, while 1 is ok.


Answer: first graph  


6 0
4 years ago
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