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nikdorinn [45]
3 years ago
13

A bag contains 12 slips of paper of the same size. Each slip has one number on it, 1-12. Describe an impossible event.

Mathematics
1 answer:
Stolb23 [73]3 years ago
6 0
There are many impossible events. Here are a few. 
I'm sure you can think of at least 15 more:

-- Pull 5 multiples of 3 from the bag.

-- Pull 3 multiples of 5 from the bag.

-- Pull 7 odd numbers from the bag.

-- Pull 7 even numbers from the bag.

-- Pull two slips from the bag whose sum is more than 21 .

-- Pull three slips from the bag whose sum is more than 33 .

-- Pull two slips from the bag whose sum is less than 3 .

-- Pull three slips from the bag whose sum is less than 6 .

-- Pull two slips from the bag whose product is more than 132 .

-- Pull two slips from the bag whose product is less than 2 .

etc.
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4. Steve has a card collection. He keeps 12 of the cards in a pocket, which
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Read 2 more answers
Given x^2 + y^2 = 36 and xy = -10 find x + y.
9966 [12]
This looks fun

ok so
we will use subsitution
xy=-10
divide both sides by x or y (I will choose y)
x=-10/y

sub -10/y fo x

(-10/y)^2+y^2=36
100/(y^2)+y^2=36
times both sides by y^2
100+y^4=36y^2
minus 36y^2 from both sides
y^4-36y^2+100=0
(y^2)^2-36(y^2)+100=0
quadratic formula
for
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}
x=\frac{-(-36)+/- \sqrt{(-36)^2-4(1)(100)} }{2(1)}
x=18+/- 4\sqrt{14}

sub
y=-10/x
y=\frac{-10}{18+/- 4\sqrt{14}}

so x+y=18+/- 4\sqrt{14}+\frac{-10}{18+/- 4\sqrt{14}}
multiply first number by \frac{18+/- 4\sqrt{14}}{18+/- 4\sqrt{14}} and add them

x+y=\frac{(18+/- 4\sqrt{14})(18+/- 4\sqrt{14})-10}{18+/- 4\sqrt{14}}
or
x+y=\frac{81+22 \sqrt{14} }{5} or \frac{81-22 \sqrt{14} }{5}




8 0
3 years ago
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