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Dominik [7]
2 years ago
9

Determine and describe the ​number​ and ​type​ of roots of the quadratic functions. Show any and all work necessary to determine

your answer. Justify your answer with appropriate reasoning. a) y=3x^2+7x+8 b) y=2x^2-7x-2 c) y=9x^2+30+25
Mathematics
1 answer:
algol132 years ago
7 0

Answer:

See explanations below

Step-by-step explanation:

The root of a quadratic equation is determined by its discriminant

D = b²-4ac

If D > 0, the roots are real and unique

If D < 0, the roots are complex

If D= 0, the roots are real and equal

a) For the quadratic equation

y=3x^2+7x+8

Since the highest degree of the equation is 2, hence the equation will have 2 roots

From the equation, a = 3,  b = 7 and c = 8

Get the discriminant

D = 7²-4(3)(8)

D = 49 - 96

D = -47

Since the discriminant value is less than 0, hence the roots are complex roots

b) For the quadratic equation

y=2x^2-7x-2

Since the highest degree of the equation is 2, hence the equation will have 2 roots

From the equation, a = 2,  b = -3 and c = -2

Get the discriminant

D = (-3)²-4(2)(-2)

D = 9 + 16

D = 25

Since the discriminant value is greater than 0, hence the roots are real and unique.

c) For the quadratic equation

y=9x^2+30x+25

Since the highest degree of the equation is 2, hence the equation will have 2 roots

From the equation, a = 9,  b = 30 and c = 25

Get the discriminant

D = 30²-4(9)(25)

D = 900 - 900

D = 0

Since the discriminant value is equal to 0, hence the roots are complex real and equal

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Step-by-step explanation:

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2 years ago
Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val
Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

A(t) = Pe^{rt}

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that P = 330

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

A(t) = Pe^{rt}

590 = 330e^{8r}

e^{8r} = \frac{590}{330}

e^{8r} = 1.79

Applying ln to both sides:

\ln{e^{8r}} = \ln{1.79}

8r = \ln{1.79}

r = \frac{\ln{1.79}}{8}

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Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

A(t) = 330e^{0.0726t}

A(10) = 330e^{0.0726*10} = 682

The value of the account in the year 2009 will be $682.

4 0
3 years ago
What is thee answer?
11111nata11111 [884]
I believe the answer is B.
Hope this helps <3
5 0
2 years ago
6a+ -12a plz explain (a=1/2) I don't know evaluating expressions
Marianna [84]

Plug in the 1/2 where the a’s are answer is -3

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3 years ago
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