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____ [38]
2 years ago
11

Please Help!!! All my points are on this.

Mathematics
1 answer:
Viktor [21]2 years ago
8 0

Answer:

y = 6 sin(2x+π/2) +4

Step-by-step explanation:

Given:

-sinusoidal function

-maximum point at (0,10)

-intersects its midline at (π/4,4)

Build the function:

y = sin x , we start with this because is a sinusoidal function

y = sin (x+ π/2), to move the maximum on the y-axis where x= 0

y = sin (2x +π/2), to move the midline from π/2 to a π/4 we need

y = 4+ sin(2x+π/2) , to move the midline from (π/4, 0) to  a (π/4, 4)

y = 4+ 6 sin(2x+π/2), to move the max at (0,10), -because the midline is at 4 and the function max at 10 we need 10-4 = 6

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Answer:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}

Step-by-step explanation:

<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}

Operating on the numerators:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-(3q^2-q-6)}{q^2+6q+5}

Removing parentheses:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-3q^2+q+6}{q^2+6q+5}

Simplifying:

\boxed{\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}}

The expression cannot be further simplified.

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UkoKoshka [18]

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Read 2 more answers
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