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NeX [460]
3 years ago
9

Can someone help me with this question please and thank you.

Mathematics
1 answer:
Ainat [17]3 years ago
3 0
Where’s the question at?
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Find the 12th term of the arithmetic sequence 2x + 8,6x + 5, 10x + 2, ...
lawyer [7]

Answer: The 12th term is 46x -25

Step-by-step explanation:

It is increasing by 4x and decreasing by 3 through each term, this represents the d. a1 has already been identified as 2x +8, so you just have to write down the formula and the solve form there by plugging in the 12 in (n-1) to find the answer.

8 0
2 years ago
Find the gradient of the line segment between the points (-3,3) and (4,4).
ElenaW [278]

Answer:

2.

Step-by-step explanation:

The gradient or slope of a straight line which is found between two points (x1, y1) and (x2, y2) is given by the formula:

m = (y2 - y1) / (x2 - x1)

Where y2 = 7

y1 = 3

x2 = 4

x1 = 2

=> m = (7 - 3) / (4 - 2)

m = 4 / 2 = 2

The gradient is 2.

4 0
3 years ago
It's number 6?its really hard me and my friends have tried but we just don't know.
Elden [556K]
Let n be the amount of points earned in round 5

the goal of the game is to get zero when you add up the point value from all the rounds.

this means
-9 +3 + (-8)+ 2 +n =0

add up all the given numbers, then see which number(n) is required in order to get zero

any questions?
6 0
2 years ago
members of students council set up 96 chairs for a school meeting. they placed the chairs in rows,with an equal number of chairs
Sauron [17]

Answer:

6

Step-by-step explanation:

96÷5=19.2

<u>96÷6=16</u>

96÷10=9.6

96÷14=6.857142~

6 0
2 years ago
Find the equation of the line that is perpendicular to the line y = (-1/3)x -1 and passes through the point (1, 5)?
Anit [1.1K]

bearing in mind that perpendicular lines have negative reciprocal slopes, so


\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill


\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{1}\implies 3}}


so we're really looking for a line whose slope is 3 and runs through (1,5)


\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{5})~\hspace{10em} slope = m\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-5=3(x-1) \\\\\\ y-5=3x-3\implies y=3x+2

4 0
3 years ago
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