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Scrat [10]
3 years ago
7

There are eight people competing in the track meet this weekend. They are competing for 1st, 2nd, and 3rd place ribbons. How man

y different ways can the winners be chosen? A. 67 B. 306 C. 336 D. 8,064
Mathematics
1 answer:
malfutka [58]3 years ago
5 0

Answer:

Option C

Step-by-step explanation:

Assuming P represents the number of outcomes, n the number of competitors and r the number of places to be awarded....

P = n! / ( n - r )!\\P = 8! / ( 8 - 3 )!\\P = 8! / 5!\\\\Algebra, P = 336 \\Solution = Option C

There are 336 different possible ways for the winners to be chosen.

<u><em>Hope that helps!</em></u>

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If f(x) = х2 – 2x and g(x) = 6х +4, for which vаluе оf x does (f+g)(x) = 0?
lana66690 [7]

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2 years ago
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4 0
3 years ago
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MrRa [10]
Check the picture below.

bear in mind that, the "bases" are the two parallel sides, and the height is the distance between them.

\bf \textit{area of this trapezoid}\\\\&#10;A=\cfrac{AB(BC+AD)}{2}\\\\&#10;-------------------------------\\\\&#10;\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ -1}}\quad ,&{{ 5}})\quad &#10;%  (c,d&#10;B&({{ 3}}\quad ,&{{ 2}})&#10;\end{array}\qquad &#10;%  distance value&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}&#10;\\\\\\&#10;AB=\sqrt{[3-(-1)]^2+[2-5]^2}\implies AB=\sqrt{(3+1)^2+(2-5)^2}&#10;\\\\\\&#10;AB=\sqrt{16+9}\implies AB=\sqrt{25}\implies \boxed{AB=5}

\bf -------------------------------\\\\&#10;\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;B&({{ 3}}\quad ,&{{ 2}})\quad &#10;%  (c,d&#10;C&({{ 0}}\quad ,&{{ -2}})&#10;\end{array}&#10;\\\\\\&#10;BC=\sqrt{(0-3)^2+(-2-2)^2}\implies BC=\sqrt{9+16}&#10;\\\\\\&#10;BC=\sqrt{25}\implies \boxed{BC=5}

\bf -------------------------------\\\\&#10;\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ -1}}\quad ,&{{ 5}})\quad &#10;%  (c,d&#10;D&({{ -13}}\quad ,&{{ -11}})&#10;\end{array}&#10;\\\\\\&#10;AD=\sqrt{[-13-(-1)]^2+[-11-5]^2}&#10;\\\\\\&#10;AD=\sqrt{(-13+1)^2+(-16)^2}\implies AD=\sqrt{144+256}&#10;\\\\\\&#10;AD=\sqrt{400}\implies \boxed{AD=\sqrt{20}}

so, the area for this trapezoid is then

\bf A=\cfrac{5(5+20)}{2}\implies A=\cfrac{125}{2}\implies A=62\frac{1}{2}

8 0
3 years ago
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