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shusha [124]
3 years ago
9

A quadratic function has a line of symmetry at x = –3.5 and a zero at –9. What is the distance from the given zero to the line o

f symmetry? What is the other zero of the quadratic function?
Mathematics
1 answer:
kap26 [50]3 years ago
8 0

Given:

A quadratic function has a line of symmetry at x = –3.5 and a zero at –9.

To find:

The other zero.

Solution:

We know that, the line of symmetry divides the graph of quadratic function in two congruent parts. So, both zeroes are equidistant from the line of symmetry.

It means, line of symmetry passes through the mid point of both zeroes.

Let the other zero be x.

-3.5=\dfrac{(-9)+x}{2}

Multiply both sides by 2.

-7=-9+x

Add 9 on both sides.

-7+9=-9+x+9

2=x

Therefore, the other zero of the quadratic function is 2.

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Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.
andriy [413]

Answer:

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Step-by-step explanation:

We have been given a indefinite integral \int \frac{2}{3x\left(3x-5\right)}dx. We are asked to find the indefinite integral.

We will use partial fraction formula to solve our given problem.

\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}

\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx

\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx

Using difference rule of integrals, we will get:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)

Now, we need to use u-substitution as:

Let u=3x-5.

\frac{du}{dx}=3

dx=\frac{1}{3}du

\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|

\int \frac{1}{5x}dx=\frac{1}{5}\int \frac{1}{x}dx=\frac{1}{5}ln|x|

Substitute back these values:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)=\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)

Let us add a constant C.

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Therefore, our required integral would be \frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C.

5 0
3 years ago
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