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AleksandrR [38]
3 years ago
7

Expand the following equation:5x(3-7y)​

Mathematics
2 answers:
densk [106]3 years ago
4 0

Answer:

15x - 35xy

5x * 3 - 5x * 7y

15x - 35xy

Leokris [45]3 years ago
3 0

Answer:

Step-by-step explanation:

15x-35x

25x is the answer

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At what rate is the temperature decreasing?
Murljashka [212]

Answer:

8 celcius

Step-by-step explanation:

The slope is -8 so it is decreasing by 8

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Find B if the graph of the equation 3x=By+2 is perpendicular to the graph of 3y=-2x+4.
Romashka [77]

Answer:

B = \frac{3x}{y} - \frac{2}{y}

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ILL GIVE YOU BRAINLIST !! * have to get it right * Find the slope of the line represented in the table.
cestrela7 [59]

Answer:

2/3 (B)

Step-by-step explanation:

The y values increase by 2 every time and the x values increase by 3. The rule for slope is rise(y)/run(x) so you get b, 2/3

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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

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3 years ago
Consider the line y=8x-7
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Answer:

1.8

2.-1/8

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Just take the number next to the x for #1 Just take the x away and for #2 just make a fraction by doing this 1/8

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2 years ago
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