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Vinil7 [7]
3 years ago
10

Write a function rule for the table. x f(x) (–3, 1) (–2, 2) (–1, 3) (0, 4)

Mathematics
2 answers:
inn [45]3 years ago
7 0
The answer is (1,1) a liner function
Mila [183]3 years ago
3 0
Hope this helps but the function rule for this table is f(x)=x+4
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What type of triangle is formed by joining the points D(7,3), E(8,1) and F(4,-1)
andreyandreev [35.5K]

Calculate the lengths of the sides of the triangle:

The triangle has three different sides, so it's a scalene triangle.

Now if c is the longest side of the triangle, and a and b are the shorter sides, then:

- if , the triangle is right

- if , the triangle is acute

- if , the triangle is obtuse

The longest side is 5.

The square of the longest side is equal to the sum of the squares of the shorter sides, so it's a right triangle.

The triangle is a right scalene triangle.

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3 years ago
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tino4ka555 [31]

Answer:

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Step-by-step explanation:

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3 years ago
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6<br> x<br> −<br> 5<br> y<br> =<br> 14<br><br> Find <br> x<br> when <br> y<br> =<br> 2
sattari [20]

Answer:

4

Step-by-step explanation:

3 0
3 years ago
Which of the following options best describes the function graphed below?
wolverine [178]
Nonlinear increasing
6 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

When r = 6

\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28

When r = 7

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

When r = 8

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

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0
3 years ago
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