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Fofino [41]
3 years ago
13

Find the discriminant to determine the number of roots, then solve the quadratic equation using the quadratic formula. !! PLS HE

LP !!
Mathematics
1 answer:
VikaD [51]3 years ago
7 0

Answer:

Discriminant review

Discriminant reviewThe discriminant is the part of the quadratic formula underneath the square root symbol: b²-4ac. The discriminant tells us whether there are two solutions, one solution, or no solutions.

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a rectangular corn hole area at the recreation center has a width of 5 feet and a length of 10 feet. if a uniform amount is adde
harkovskaia [24]
Let
x-------------> amount added <span>to each side
</span>A-------------> area increased----------> 84 ft²

we know that
A=(10+x)*(5+x)=84-----------> 10*5+10*x+5*x+x²
50+15x+x²=84----------> x²+15x-34=0
<span>solving the second order equation
</span>x1=-17
x2=2

 the answer is x=2 ft
3 0
3 years ago
Read 2 more answers
A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow
Mrrafil [7]

Answer:

\displaystyle \frac{54}{5405}.

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

\displaystyle \left(50\atop 5\right) = 2,118,760.

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

\displaystyle \left( 28\atop 2\right) = 378.

Number of ways to choose 3 green candies out of a batch of 8:

\displaystyle \left(8\atop 3\right)=56.

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168.

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}.

3 0
3 years ago
I need help with the answers
vlada-n [284]

Answer:

i dont get it

Step-by-step explanation:

4 0
3 years ago
A box contains 3 yellow, 2 red, 4 green and 3 black marbles. Two marbles are taken one after the other at random from the box. W
Nata [24]
I believe it's 2/10.
7 0
3 years ago
I need some help with this question..
Anuta_ua [19.1K]
It is (B) I think:) hope this helps
6 0
3 years ago
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