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3 years ago

Answers................

Mathematics

1 answer:

Andreyy893 years ago

6 0

Answer:

x=1

Step-by-step explanation:

log2( x^2 -x+2) = 1+2log2(x)

Rewriting 1 as log2(2)

log2( x^2 -x+2) = log2(2)+2log2(x)

We know that a log b = log a^b

log2( x^2 -x+2) = log2(2)+log2(x^2)

we know log a + log b = log (ab)

log2( x^2 -x+2) = log2(2*x^2)

Since the bases are the same the terms inside must be equal

x^2 -x+2 = 2x^2

Subtract 2x^2 from each side

-x^2 -x+2 = 0

Multiply by -1

x^2 +x-2 = 0

Factor

(x+2)(x-1)=0

Using the zero product property

x+2 = 0 x-1=0

x=-2 x=1

Checking the solutions

log2( x^2 -x+2) = 1+2log2(x)

X cannot be negative because 2 log2(x) cannot be negative

log2( 1^2 -1+2) = 1+2log2(1)

x=1

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2 years ago

What is PA Enter your answer in the box

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Step-by-step explanation:

P is the in-center

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3 years ago

Quadratics need help making parabola

timama [110]

Answer:

The equation of the parabola is $y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4$ , whose real vertex is $(x,y) = (2, -5.333)$ , not $(x,y) = (2, -5)$ .

Step-by-step explanation:

A parabola is a second order polynomial. By Fundamental Theorem of Algebra we know that a second order polynomial can be formed when three distinct points are known. From statement we have the following information:

$(x_{1}, y_{1}) = (-2, 0)$ , $(x_{2}, y_{2}) = (6, 0)$ , $(x_{3}, y_{3}) = (0, -4)$

From definition of second order polynomial and the three points described above, we have the following system of linear equations:

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$36\cdot a + 6\cdot b + c = 0$ (2)

$c = -4$ (3)

The solution of this system is: $a = \frac{1}{3}$ , $b = - \frac{4}{3}$ , $c = -4$ . Hence, the equation of the parabola is $y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4$ . Lastly, we must check if $(x,y) = (2, -5)$ belongs to the function. If we know that $x = 2$ , then the value of $y$ is:

$y = \frac{1}{3}\cdot (2)^{2}-\frac{4}{3}\cdot (2) - 4$

$y = -5.333$

$(x,y) = (2, -5)$ does not belong to the function, the real point is $(x,y) = (2, -5.333)$ .

5 0

2 years ago

Answers................ ​

Answers................