3 years ago

Match each equation with its number of unique solutions.

Mathematics

1 answer:

Anni [7]3 years ago

4 0

Answer:

y = 3x^2-6x+3 one real solution

y = -x^2 - 4x + 7 two real solution

y = -2x^2+9x-11 two complex solutions

Step-by-step explanation:

b^2-4ac = 0 1 repeated real solution

b^2-4ac > 0 2 distinct real solutions

b^2-4ac < 0 2 complex solutions

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Mr. frankel bought 7 tickets to a puppet show and spent $24. he bought a combination of child tickets for $2 each and adult tick

nydimaria [60]

Let x be the number of child tickets he bought
Let y be the number of adult tickers he bought

① x+y=7 (child tickets+adult ticket=7 tickets in total)
② 2x+4y=24 (price of child tickets+price of adult tickets=$24 in total)

We may simply the second equation since all of the coefficients are divisible by 2.

① x+y=7
② x+2y=12

We can now use elimination by multiplying the second equation by -1.

② -(x+2y=12)
② -x-2y=-12

① x+y=7
② -x-2y=-12

Now putting the equations together,
-y=-5
y=5
x=2

Therefore he bought 2 child tickets and 5 adult tickets

6 0

4 years ago

PLZZZ HELP!!! VERY CONFUSED!!! Will give brainliest!!!!

777dan777 [17]

By the intersecting chords theorem, you have

$7\cdot4=5\cdot x\implies 5x=28\implies\boxed{x=\dfrac{28}5}$