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inysia [295]
2 years ago
11

Marco went bowling and spent a total of $21. This included the shoe rental of $7. He bowled 4 games. How much did each game cost

? A. $3.50 B. $3.66 C. $4.25 D. $7.00
Mathematics
2 answers:
Bas_tet [7]2 years ago
4 0
The costs of each game was A. $3.50
$21 - $7\4 = $3.50

IRISSAK [1]2 years ago
3 0
So first you do 21 - 7 because you don't want to count the shoe rental, and you get 14. Then, if he bowled 4 games, you'd divide 14 by 4 which is (1 ÷ 4 = 0
14 ÷ 4 = 3 (remainder 2)
20 ÷ 4 = 5)
3.5, which is A. $3.50
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I really need help with this problem!!! Can someone help!!!! Please!!!
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  (a)  31.5 in²

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3 years ago
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vaieri [72.5K]

Answer:

see explanation

Step-by-step explanation:

Note the common difference d between consecutive terms of the sequence

8 - 6 = 10 - 8 = 2

This indicates that the sequence is arithmetic with n th term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 6 and d = 2, thus

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8 0
3 years ago
What is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through the
Svetradugi [14.3K]

keeping in mind that perpendicular lines have negative reciprocal slopes, hmmmm what's the slope of that line above anyway,

\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{2+1}{3}\implies 1 \\\\[-0.35em] ~\dotfill

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

so we're really looking for the equation of a line whose slope is -1 and runs through (2,5)

\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{5})~\hspace{10em} \stackrel{slope}{m}\implies -1 \\\\\\ \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-\stackrel{y_1}{5}=\stackrel{m}{-1}(x-\stackrel{x_1}{2}) \\\\\\ y-5=-x+2\implies y=-x+7

8 0
3 years ago
Read 2 more answers
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