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2 years ago

Y^2 - 12y + 27 / y^2 - 6y - 27 simply the rational equation.

Mathematics

1 answer:

Ulleksa [173]2 years ago

3 0

The simplified rational expression is (y - 3)/(y + 3). Where y ≠ -3.

<h3>How to simplify a rational expression?</h3>

A rational expression is in the p/q form. Where p and q are polynomial functions.

To simplify this rational equation,

Factorize the polynomials in both numerator and denomiantor.
Cancel out common factors if any.
If the denominator and the numerator have no common factors except 1, then that is said to be the simplest form of the given rational expression.

<h3>Calculation:</h3>

The given rational equation is

$\frac{y^2 - 12y + 27 }{y^2 - 6y - 27}$

Factorizing the expression in the numerator:

y² - 12y + 27 = y² - 9y - 3y + 27

⇒ y(y - 9) - 3(y - 9)

⇒ (y - 3)(y - 9)

Factorizing the expression in the denominator:

y² - 6y - 27 = y² - 9y + 3y - 27

⇒ y(y - 9) + 3(y - 9)

⇒ (y + 3)(y - 9)

Since they have (y - 9) as the common factor, we can simplify,

$\frac{y^2 - 12y + 27 }{y^2 - 6y - 27}=\frac{(y-3)(y-9)}{(y+3)(y-9)}$

⇒ (y - 3)/(y + 3) where y ≠ -3(denomiantor)

Here there are no more common factors except 1; this is the simplest form of the given rational expression.

Learn more about simplifying rational expressions here:

brainly.com/question/1928496

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Is there statistically significant evidence that the districts with smaller classes have higher average test scores? The t-sta

Musya8 [376]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The 95% confidence interval is $[670.03 , 673.97 ]$

The test statistics is $t = 7.7$

The p-value is $p-value = 0$

The <u>p-value</u> suggests that the null hypothesis is<u> rejected </u>with a high degree of confidence. Hence <u>there is</u> statistically significant evidence that the districts with smaller classes have higher average test score

Step-by-step explanation:

From the question we are told that

The sample size is n = 408

The sample mean is $\= y = 672.0$

The standard deviation is $s = 20.3$

Given that the confidence level is 95% then the level of significance is

$\alpha = (100 -95 )\% = 0.05$

From the normal distribution table the critical value of $\frac{\alpha }{2} = \frac{0.05 }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }$

=> $E = 1.96 * \frac{20.3}{\sqrt{408} }$

=> $E = 1.970$

Generally the 95% confidence interval is mathematically represented as

$\= y -E < \mu < \= y + E$

=> $672.0 -1.970 < \mu < 672.0 +1.970$

=> $670.03 < \mu < 673.97$

=> $[670.03 , 673.97 ]$

From the question we are told that

Class size small large

$Avg.score(\= y)$ $\= y_1 = 683.7$ $\= y_2 = 676.0$

$S_y$ $S_{y_1} =20.2$ $S_{y_2} = 18.6$

sample size $n_1 = 229$ $n_2 = 184$

The null hypothesis is $H_o : \mu_1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu_1 - \mu_2 > 0$

Generally the standard error for the difference in mean is mathematically represented as

$SE = \sqrt{\frac{S_{y_1}^2 }{n_1} +\frac{S_{y_2}^2 }{n_2} }$

=> $SE = \sqrt{20.2^2 }{229} +\frac{18.6^2 }{184_2} }$

=> $SE = 1.913$

Generally the test statistics is mathematically represented as

$t = \frac{\= y _1 - \= y_2 }{SE}$

=> $t = \frac{683.7 - 676.0 }{1.913}$

=> $t = 7.7$

Generally the p-value is mathematically represented as

$p-value = P(t > 7.7 )$

From the z-table

$P(t > 7.7 ) = 0$

$p-value = 0$

From the values we obtained and calculated we can see that $p-value < \alpha$

This mean that

The p-value suggests that the null hypothesis is rejected with a high degree of confidence. Hence there is statistically significant evidence that the districts with smaller classes have higher average test score

4 0

3 years ago

Use the properties of real numbers to rewrite the expression. 2/5 * (-7) * 5/2

Mama L [17]

For this case we have the following expression:

$\frac{2}{5}*(- 7)*\frac{5}{2}$

Using the associative property we can rewrite the expression in the following way:

$(\frac{2}{5}*\frac{5}{2})*(-7)$

Finally, simplifying we have:

$(1) (- 7)$

$-7$

Answer:

Rewriting the expression we have that the result is:

d. -7

4 0

3 years ago

A track-and-field athlete releases a javelin. The height of the javelin as a function of time is shown on the graph below. Use t

Georgia [21]

Part 1:

For this case we must see in the graph the axis of symmetry of the given parabola.
We have then that the axis of symmetry is the vertical line t = 2.
Answer:
The height of the javelin above the ground is symmetric about the line t = 2 seconds:

Part 2:

For this case, we must see the time t for which the javelin reaches a height of 20 feet for the first time.
We then have that when evaluating t = 1, the function is h (1) = 20. To do this, just look at the graph.
Then, we must observe the moment when it returns to be 20 feet above the ground.
For this, observing the graph we see that:
h (3) = 20 feet
Therefore, a height of 20 feet is again reached in 3 seconds.
Answer:
The javelin is 20 feet above the ground for the first time at t = 1 second and again at t = 3 seconds

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3 years ago

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Tema [17]

Answer:

Step-by-step explanation:

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