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ddd [48]
2 years ago
10

Write a polynomial of degree 3 whose only roots are x=2 and x=2/5. Is there another polynomial of degree 3 that has the same roo

ts?
Mathematics
2 answers:
TEA [102]2 years ago
7 0
One such polynomial might be

(x-2)\left(x-\dfrac25\right)^2

Another with the same roots would be

(x-2)^2\left(x-\dfrac25\right)
Allushta [10]2 years ago
5 0

Answer:  The answer is 5x^3-22x^2+28x-8 and 25x^3-70x^244x-8.

Step-by-step explanation:  We are given to write a polynomial of degree 3 whose only roots are 2 and \dfrac{2}{5}. Also, we need to find another polynomial of degree 3, if exists, that has the same roots.

The polynomial will be given by

p(x)=(x-2)(x-2)(x-\dfrac{2}{5})\\\\\Rightarrow 5p(x)=(x^2-4x+4)(5x-2)\\\\\Rightarrow P(x)=5x^3-20x^2+20x-2x^2+8x-8\\\\\Rightarrow P(x)=5x^3-22x^2+28x-8.

The other polynomial with same roots exist and is given by

q(x)=(x-2)(x-\dfrac{2}{5})(x-\dfrac{2}{5})\\\\\Rightarrow 25q(x)=(x-2)(5x-2)^2\\\\\Rightarrow Q(x)=(x-2)(25x^2-20x+4)\\\\\Rightarrow Q(x)=25x^3-20x^2+4x-50x^2+40x-8\\\\\Rightarrow Q(x)=25x^3-70x^244x-8.

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Which expression is equivalent to
nirvana33 [79]

For this case, we must find an expression equivalent to:

\frac {-9x ^ {- 1} y ^ {- 9}} {- 15x ^ 5y ^ {- 3}}

By definition of power properties we have:

a ^ {- 1} = \frac {1} {a ^ 1} = \frac {1} {a}

Rewriting the previous expression we have:

The "-" are canceled and we take into account that:

\frac {9} {15} = \frac {3} {5}

So:

\frac {3} {5x^5 * x ^ 1 *y ^ 9* y ^ {- 3}} =

According to one of the properties of powers of the same base, we must put the same base and add the exponents:

\frac {3} {5x ^ {5 + 1} * y ^ {9-3}} =\\\frac {3} {5x ^ 6 * y ^ 6}

Answer:

\frac {3} {5x ^ 6 * y ^ 6}

Option B

3 0
3 years ago
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The graph of the function f(x) = x2 + 8x + 12 is shown. Which statements describe the graph? Check all that apply.
strojnjashka [21]
<span>The axis of symmetry is x = –4.
</span><span>The domain is all real numbers.
</span><span>The x-intercepts are at (–6, 0) and (–2, 0). </span>
7 0
3 years ago
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The credit remaining on a phone card (in dollors) is a linear function of the total calling trime made with the card ( in minute
KATRIN_1 [288]

After 85 minutes of calls, there are $27.25 left on the card.

<h3>What is the remaining credit after 85 minutes of calls?</h3>

A linear equation in slope-intercept form is:

y = a*x + b

Where a is the slope.

If the line passes through two points (x₁, y₁) and (x₂, y₂) the slope is:

a = \frac{y_2 - y_1}{x_2 - x_1}

In this case, we know that the line passes through the points (22, 36.7) and (52, 32,20)

(points of the form (time, dollars)).

So the slope is:

a = \frac{32.2 - 36.7}{52 -22} = -0.15

The linear equation is then:

y = -0.15*x + b

To find the value of b, we use the point (22. 36.7)

36.7 = -0.15*22 + b

36.7 + 0.15*22 = b = 40

Then the linear equation is:

y = -0.15*x +40

The amount remaining in the credit card after 85 minutes is given by evaluating the above equation in x = 85.

y = -0.15*85 + 40 = 27.25

This means that after 85 minutes of calls, there are $27.25 left on the card.

If you want to learn more about linear equations:

brainly.com/question/1884491

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7 0
1 year ago
Suppose f and g are continuous functions such that g(6) = 6 and lim x → 6 [3f(x) + f(x)g(x)] = 45. Find f(6).
aleksandr82 [10.1K]
Since g(6)=6, and both functions are continuous, we have:

\lim_{x \to 6} [3f(x)+f(x)g(x)] = 45\\\\\lim_{x \to 6} [3f(x)+6f(x)] = 45\\\\lim_{x \to 6} [9f(x)] = 45\\\\9\cdot lim_{x \to 6} f(x) = 45\\\\lim_{x \to 6} f(x)=5


if a function is continuous at a point c, then lim_{x \to c} f(x)=f(c), 

that is, in a    c ∈  a continuous interval, f(c) and the limit of f as x approaches c are the same.


Thus, since lim_{x \to 6} f(x)=5, f(6) = 5


Answer: 5


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3 years ago
Which expression is equivalent to 4(-3h+2) - (h-8) ???
Lera25 [3.4K]

You have to use distributive property then you combine like terms

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