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4 years ago

How to classify this conic ecwation 4x^2+9y^2-54y-63=0

1 answer:

6 0

This looks like an ellipse.
Complete the square for y: 9(y²-6y-7)=9(y²-6y+9-9-7)=9((y-3)²-16).
Rewrite the equation:
4x²+9(y-3)²-144=0; 4x²/144+9(y-3)²/144=1; x²/36+(y-3)²/16=1.
Standard equation of an ellipse is (x-h)²/a²+(y-k)²/b²=1, where (h,k) is the origin or centre.
This is the equation of an ellipse with centre (0,3), a=6 and b=4 (semimajor and semiminor axes).

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Find the mean of the data in the dot plot below.

Nataly [62]

Answer:

5km

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Having A really hard time and my Professor is being a A Hole with help so can someone just solve it for me I need the range and

jeka57 [31]

Normally, when $x=3$ the function

$f(x)=\dfrac12(x-1)^3-1$

takes on a value of

$f(3)=\dfrac12(3-1)^3-1=\dfrac82-1=4-1=3$

so the graph of $f(x)$ passes through the point (3, 3). This is the only place at which the cubic takes on a value of 3, since $f(x)$ is one-to-one.

In the definition of $q(x)$ , however, we remove this point from the curve and replace with the point (3, -2).

While $f(x)$ is a standard cubic function, so that its range is all real numbers, the replacement made for $q(x)$ removes 3 from the range set, so that the range of $q(x)$ would be all real numbers *except* 3.

3 0

4 years ago

Calvin buys a book for $14.70, which is a %30 discount off the original price

tatyana61 [14]

$10.29 (14.7)(.3)= 4.41. 14.7-4.41

8 0

3 years ago

A heated piece of metal cools according to the function c(x) = (.5)^x − 11, where x is measured in hours. A device is added that

Goryan [66]

Answer:

The temperature of metal after five years = -18.96875

Step-by-step explanation:

Given : Function $c(x) = (0.5)^x-11$

where x is measured in hours

Another function is added h(x)=-x-3

therefore, the resultant function f(x)=h(x)+c(x)

$f(x) = (0.5)^x-11-x-3$

For x=5

$f(5) = (0.5)^5-11-5-3$

$f(5) = (0.03125)-19$

$f(5) = -18.96875$

Therefore, The temperature of metal after five years = -18.96875

5 0

3 years ago

Read 2 more answers

A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise

DiKsa [7]

Answer:

We Reject H₀ if t calculated > t tabulated

But in this case,

0.83 is not greater than 2.056

Therefore, we failed to reject H₀

There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.

Step-by-step explanation:

Refer to the attached data.

The Null and Alternate hypothesis is given by

Null hypotheses = H₀: μ₁ = μ₂

Alternate hypotheses = H₁: μ₁ ≠ μ₂

The test statistic is given by

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } }$

Where $\bar{x}_1$ is the sample mean of people who do not exercise regularly.

Where $\bar{x}_2$ is the sample mean of people who do exercise regularly.

Where $s_1$ is the sample standard deviation of people who do not exercise regularly.

Where $s_2$ is the sample standard deviation of people who do exercise regularly.

Where $n_1$ is the sample size of people who do not exercise regularly.

Where $n_2$ is the sample size of people who do exercise regularly.

$t = \frac{72.7 - 69.7}{\sqrt{\frac{10.9^2}{16} + \frac{8.2^2}{12} } }$

$t = 0.83$

The given level of significance is

1 - 0.95 = 0.05

The degree of freedom is

df = 16 + 12 - 2 = 26

From the t-table, df = 26 and significance level 0.05,

t = 2.056 (two-tailed)

Conclusion:

We Reject H₀ if t calculated > t tabulated

But in this case,

0.83 is not greater than 2.056

Therefore, We failed to reject H₀

There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.

4 0

3 years ago