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Ira Lisetskai [31]
3 years ago
11

Answer question 1 and 2 on the champions challenge? it really matters

Mathematics
1 answer:
babunello [35]3 years ago
8 0
If length = a, width = b
if double both a and b will double its area?
area of a rectangle is ab
if double both a and b then the area is (2a)(2b) = 4ab (then peter is wrong because the area will be 4 times not double)
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Multiply (x−3)(x+2)
Phantasy [73]

We FOIL, multiply firsts, outers, inners and lasts, and add them all up.

(x-3)(x+2) = x·x + 2x - 3x + (-3)(2) = x² - x - 6

Answer: x² - x - 6

4 0
4 years ago
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Which of the following is NOT an equation?
zavuch27 [327]
Number 2 is not an equation.
3 0
3 years ago
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Assume that y varies directly with x then solve y = 2.5 x = .5 when x = 20
velikii [3]

Step-by-step explanation:

y=2.5| ?

x=0.5| 20

It'll be criss-crossed then it'll be:-

2.5×20=50

5 0
3 years ago
What is the probability of flipping a coin 6 times and getting all heads
abruzzese [7]
The probability of getting all heads is 1 / 2^6 = 1/64 as there is only 1 event where this happens in a possible 2^6 = 64 events. It is the same as the probability of getting all tails. The probability of getting at least 1 head is 1 - p(all tails) = 63/64.
7 0
2 years ago
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I NEED HELP ON THIS PROBLEM!!! PLEASE
Allushta [10]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

5x+8y=-9\implies 8y=-5x-9\implies y=\cfrac{-5x-9}{8} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{5}{8}} x-9\qquad \impliedby \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}

well then, so since this equation has that slope therefore

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{-5}{8}} ~\hfill \stackrel{reciprocal}{\cfrac{8}{-5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{8}{-5}\implies \cfrac{8}{5}}}

so we're really looking for the equation of a line whose slope is 8/5 and runs through (10,10)

(\stackrel{x_1}{10}~,~\stackrel{y_1}{10}) ~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{8}{5} \\\\\\ \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}{10}=\stackrel{m}{\cfrac{8}{5}}(x-\stackrel{x_1}{10})

5 0
3 years ago
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