Which of the following expressions are equivalent to 6-3? DONE

The center of an ellipse is (−9,3) . One focus is (−13,3) . The major axis is 18 units long. What is the equation of the ellipse

Montano1993 [528]

The equation of the ellipse in standard form is:

$\frac{(x+9)^{2}}{81}+\frac{(y-3)^{2}}{65}=1$

Step-by-step explanation:

The standard form of the equation of an ellipse with center (h , k)

and major axis parallel to x-axis is $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$

1. The length of the major axis is 2a

2. The coordinates of the vertices are (h ± a , k )

3. The coordinates of the foci are (h ± c , k), where c² = a² - b²

∵ The center of the ellipse is (-9 , 3)

∴ h = -9 , k = 3

∵ One focus is (-13 , 3)

∴ Major axis parallel to x-axis

∵ The length of the major axis = 2a

∵ The length of the major axis = 18

∴ 2a = 18

- Divide both sides by 2

∴ a = 9

∵ The coordinates of the foci are (h ± c , k)

∵ One focus is (-13 , 3)

∴ h ± c = -13

∵ h = -9

∴ -9 - c = -13

- Add 9 to both sides

∴ -c = -4

- Divide both sides by -1

∴ c = 4

∵ c² = a² - b²

∵ c = 4 and a = 9

∴ (4)² = (9)² - b²

∴ 16 = 81 - b²

- Subtract 81 from both sides

∴ -65 = -b²

- Divide both sides by -1

∴ b² = 65

- Substitute the values of a² and b² in the equation of the ellipse

∵ a² = (9)² = 81

∵ b² = 65

∵ h = -9 and k = 3

∵ $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$

- Substitute the values in the equation

∴ $\frac{(x+9)^{2}}{81}+\frac{(y-3)^{2}}{65}=1$

The equation of the ellipse in standard form is:

$\frac{(x+9)^{2}}{81}+\frac{(y-3)^{2}}{65}=1$

Learn more:

You can learn more about equations of conics equations in brainly.com/question/3333996

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3 years ago

It said I would level up from Ambitious on brainly, and I haven't. It said I would when I get 5 brainliest, and 500 points. But,

boyakko [2]

Sometimes it takes awhile before you get a new rank so, just be a little patient

6 0

3 years ago

Read 2 more answers

The difference of two numbers is 7. Three times the greater number is 72. find the numbers

atroni [7]

Answer:

$\huge\boxed{\tt{x = 24, y = 17}}$

Step-by-step explanation:

Let the greater number be x and the smaller number be y.

So,

Condition # 1:

x - y = 7 --------------(1)

Condition # 2:

3x = 72 --------------(2)

Taking Eq. (2)

3x = 72

Divide both sides by 3

x = 72 / 3

x = 24

Now, Put x = 24 in Eq. (1)

24 - y = 7

Subtract 24 to both sides

-y = 7 - 24

-y = -17

y = 17

$\rule[225]{225}{2}$

Hope this helped!

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3 years ago

Gina looks at the architectural plan of a four-walled room in which the walls meet each other at right angles. The length of one

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No. A square has equal sides that are perpendicular and parallel. I'm guessing this room resembles some other figure.

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3 years ago

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A college entrance exam company determined that a score of 2323 on the mathematics portion of the exam suggests that a student i

aleksandrvk [35]

Answer:

a) The null and alternative hypothesis are:

$H_0: \mu=23\\\\H_a:\mu> 23$

c) Test statistic t=1.53

P-value=0.064

d) The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that students who complete the core curriculum are ready for college-level mathematics. That is that the true score for the group is not significantly higher than 23.

Step-by-step explanation:

The question is incomplete:

a) State the appropriate null and alternative hypotheses.

c) Use the P-value approach at the 0.05 level of significance to test the hypotheses in part (a). ldentify the test statistic. (Round to two decimal places as needed.) Identfy the P-value. P-value (Round to three decimal places as needed.)

d) Write a conclusion based on the results. Choose the correct answer below. ? the null hypothesis and claim that there ? sufficient evidence to conclude that the population mean is ? than 20.

This is a hypothesis test for the population mean.

The claim is that students who complete the core curriculum are ready for college-level mathematics.

Then, the null and alternative hypothesis are:

$H_0: \mu=23\\\\H_a:\mu> 23$

The significance level is assumed to be 0.05.

The sample has a size n=150.

The sample mean is M=23.4.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=3.2.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{3.2}{\sqrt{150}}=0.261$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{23.4-23}{0.261}=\dfrac{0.4}{0.261}=1.531$

The degrees of freedom for this sample size are:

$df=n-1=150-1=149$

This test is a right-tailed test, with 149 degrees of freedom and t=1.531, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>1.531)=0.064$

As the P-value (0.064) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that students who complete the core curriculum are ready for college-level mathematics.

6 0

3 years ago