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3 years ago

Graph the first six terms of a sequence where a1 = −10 and d = 3.

Mathematics

1 answer:

Neko [114]3 years ago

8 0

notice the first term is -10, and the common difference is 3, so we add 3 to get the next term.

$\bf \begin{array}{cccll}\stackrel{x}{n}&term&\stackrel{y}{value}\\\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\1&a_1&-10\\\\2&a_2&\stackrel{-10+3}{-7}\\\\3&a_3&\stackrel{-7+3}{-4}\\\\4&a_4&\stackrel{-4+3}{-1}\\\\5&a_5&\stackrel{-1+3}{2}\\\\6&a_6&\stackrel{2+3}{5}\end{array}$

check the picture below.

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Solve: <br><br> 4(n-7) -2 (n+3)-15n

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Answer:

solved

Step-by-step explanation:

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2 years ago

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2 years ago

What is the equation of this graphed line?

barxatty [35]

Answer:

$y=-\frac{1}{3}x-5$

Step-by-step explanation:

Given:

The two points on the line are:

$(x_1,y_1)=(-6,-3)\\(x_2,y_2)=(6,-7)$

Now, for a line with two points on it, the slope of the line is given as:

$m=\frac{y_2-y_1}{x_2-x_1}$

For the points $(x_1,y_1)=(-6,-3)\ and\ (x_2,y_2)=(6,-7)$ , slope is:

$m=\frac{-7-(-3)}{6-(-6)}\\m=\frac{-7+3}{6+6}\\m=\frac{-4}{12}=-\frac{1}{3}$

Now, for a line with slope 'm' and a point $(x_1,y_1)$ on it is given as:

$y-y_1=m(x-x_1)$

Plug in all the values and determine the equation of the line. This gives,

$y-(-3)=-\frac{1}{3}(x-(-6))\\\\y+3=-\frac{1}{3}(x+6)\\\\y+3=-\frac{1}{3}x-\frac{1}{3}\times 6\\\\y+3=-\frac{1}{3}x-2\\\\y=-\frac{1}{3}x-2-3\\\\y=-\frac{1}{3}x-5$

Therefore, the equation of the line is:

$y=-\frac{1}{3}x-5$

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3 years ago

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Which of the following graphs best represents the solution to the pair of equations below? y = −2x + 2 y = 2x + 2 A coordinate p

mariarad [96]

Answer:

Step-by-step explanation:

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