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2 years ago

Find the equation of a line that is parallel to the line x = - 18 and fontaine the point (-3,5)

Mathematics

1 answer:

Oksi-84 [34.3K]2 years ago

7 0

Answer:

the line defined by the equation x = -3 is parallel to x = -18, and contains the point (-3,5)

Step-by-step explanation:

Notice that this is a vertical line that goes crosses the x axis at x= -18.

A line parallel to this will also be of the form x = "constant value", meaning that all points in the line must have a fixed x-value. If the line has to go through the point (-3, 5), that that fixed x-value must be "-3".

Therefore the line defined by the equation x = -3 is parallel to x = -18, and contains the point (-3,5)

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There are 25 town homes in the Alief Community subdivision.

vodka [1.7K]

Answer:

Step-by-step explanation:

Fist, you would see that company a Provides water to 5 people, since 1/5 of 25 is 5. 25/15X9=15. 15+5=20. So, 25-20=5.

7 0

2 years ago

Industrial chemicals are stored in 55-gallon drums. The radius of the drum is 12 inches, and the height is 32 inches. what is th

Drupady [299]

Answer:

The surface area of the drum is 3317.5218 square inches.

Step-by-step explanation:

Drum's have a cylindrical form, therefore in order to calculate its surface area we need to apply the correct formula, as shown below:

$area = 2*\pi*r^2 + 2*\pi*r*h$

Where the first term of the sum is the area of the lid and bottom of the drum and the second term is the area of the walls of the drum, r is the radius and h is the height. Applying the data from the problem, we have:

$area = 2*\pi*12^2 +2*\pi*12*32\\\\area = 3317.5218 \text{ square inches}$

The surface area of the drum is 3317.5218 square inches.

4 0

3 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$