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3 years ago

Find the equation of a line through (2,-4) and (-7,-4)

1 answer:

3 0

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Find slope :
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$\text {Slope = } \dfrac{-4 - (-4)}{-7-2} = \dfrac{0}{-9} = 0$

When the slope is 0, the graph is a horizontal line.

Equation of the graph is y = -4

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Answer: y = -4
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The slope of the function on the left is multiplied by p, and is added to the y-Intercept to arrive at the function on the right

mariarad [96]

Answer:

Step-by-step explanation:

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3 years ago

Plssss Help!! Determine the horizontal asymptote for the rational function.

schepotkina [342]

I am pretty sure it’s y=1

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3 years ago

Your professor wants to determine if her students today are performing better in statistics than 10 years ago. End of semester g

shusha [124]

Answer:

z = -3

Step-by-step explanation:

If we had two independent samples, the variance is known and $n_1$ and $n_2$ are bigger than 30, the test statistic for the difference between the two population means is calculated as:

$z=\frac{x_1-x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2} } }$

Where $x_1$ and $x_2$ are the means of the samples, $s_1^2$ and $s_2^2$ are the variances of the samples and $n_1$ and $n_2$ are the size of the sample.

Finally, replacing $x_1$ by 82, $x_2$ by 88, $s_1^2$ by 112.5 , $s_2^2$ by 54, $n_1$ by 45 and $n_2$ by 36, we get that the statistic is equal to:

$z=\frac{82-88}{\sqrt{\frac{112.5}{45}+\frac{54}{36} } }=-3$

6 0

3 years ago

Markuo's window store made a mosaic for the community center. The mosaic had a 7 × 7 array of different color square tiles. If e

djverab [1.8K]

The correct value of the length of each square tile is 3½ feet long.

Answer:

Total area of mosaic = 600.25 sq.ft

Step-by-step explanation:

We are told the length of each square tile = 3½ ft.

Area of square = L²

Where L is length of side.

Thus;

Area = 3½ × 3½

Area = 49/4 sq.ft

Now, an array of 7 x 7 means 7 rows and 7 columns

Thus, total number of square tiles = 7 x 7 = 49 tiles

So,

total area of whole mosaic = number of tiles × Area of 1 tile

Total area = 49 × 49/4 = 600.25 sq.ft

7 0

3 years ago

Let D be the region bounded by the paraboloids; z = 6 - x² - y² and z = x² + y².

Liono4ka [1.6K]

Answer:

∫∫∫1 dV=4\sqrt{3}π

Step-by-step explanation:

From Exercise we have

z=6-x^{2}-y^{2}

z=x^{2}+y^{2}

we get

2z=6

z=3

x^{2}+y^{2}=3

We use the polar coordinates, we get

x=r cosθ

y=r sinθ

x^{2}+y^{2}&=r^{2}

r^{2}=3

We get at the limits of the variables that well need for our integral

x^{2}+y^{2}≤z≤3

0≤r ≤\sqrt{3}

0≤θ≤2π

Therefore, we get a triple integral

\int \int \int 1\, dV&=\int \int \left(\int_{x^2+y^2}^{3} 1\, dz\right) dA

=\int \int \left(z|_{x^2+y^2}^{3} \right) dA

=\int \int\ \left(3-(x^2+y^2) \right) dA

=\int \int\ \left(3-r^2 \right) dA

=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} (3-r^2) dr dθ

=3\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} 1 dr dθ-\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^2 dr dθ

=3\int_{0}^{2\pi} r|_{0}^{\sqrt{3}} dθ-\int_{0}^{2\pi} \frac{r^3}{3}|_{0}^{\sqrt{3}}dθ

=3\sqrt{3}\int_{0}^{2\pi} 1 dθ-\sqrt{3}\int_{0}^{2\pi} 1 dθ

=3\sqrt{3} ·2π-\sqrt{3}·2π

=4\sqrt{3}π

We get

∫∫∫1 dV=4\sqrt{3}π

7 0

3 years ago