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8_murik_8 [283]
3 years ago
7

For problems 29 - 31 the graph of a quadratic function y=ax^2 + bx + c is shown. Tell whether the discriminant of ax^2 + bx + c

= 0 is positive, negative, or zero.

Mathematics
1 answer:
Elan Coil [88]3 years ago
3 0

Answer:

29) discriminant  is positive

30) discriminant  is 0

31) discriminant  is negative

Step-by-step explanation:

the graph of a quadratic function y=ax^2 + bx + c is shown. Tell whether the discriminant of ax^2 + bx + c = 0 is positive, negative, or zero.

In the graph of question number 29 we can see that the graph intersects the x axis at two points

so the equation has 2 solutions.

When the equation has two solution then the discriminant is positive

In the graph of question number 30 we can see that the graph intersects the x axis at only one point

so the equation has only  1 solution.

When the equation has only one solution then the discriminant is equal to 0

In the graph of question number 30 we can see that the graph does not intersects the x axis  

so the equation has 2 imaginary solutions.

When the equation has two imaginary solutions then the discriminant is negative

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If x varies directly with y and x = 2.5 when y = 10, funny x when y = 16.
amid [387]

Answer:

Should be 4.0(Not sure tho. my thought process is complicated but to me it makes sense)

Step-by-step explanation:

If y=10 and x =2.5, if y were to = 20, then x would equal 5. so you take that ratio, and say 6 is 3/5's the way to 10, so you also divide 2.5 by 3/5

you get 1.5, you add that to x already and you get 4.0

so if y=16, x should =4.0

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3 years ago
Given the two expressions shown below:
kobusy [5.1K]
<h2>Answer: A. Both are rational.</h2>

Step-by-step explanation:

Although both expressions have square roots, the result of each square root is an integer, which can be expressed as a fraction.

In this sense:

Rational numbers are all numbers that can be represented as the quotient (division) of two integer numbers. This means they can be represented as a fraction in which the denominator is nonzero.

If we solve both expressions, we will be able to see that the result is an integer that can be expressed as a fraction with two integers:

\sqrt{4} + \sqrt{25}=2 + 5= 7 The result is an integer

7=\frac{7}{1}

\sqrt{4} + \sqrt{9}= 2 + 3=5 The result is an integer

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3 years ago
What is the length of the missing side if<br>Side A: 8in <br>Side B: ?<br>Side C: 14 in ​
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Answer:

side b= 11.489

Step-by-step explanation:

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24. A metal coin is located at (18, 10). How much closer or farther from Ann is the coin than the jewelry?​
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Answer:

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Step-by-step explanation:

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2 years ago
Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decrea
maw [93]

Answer:

a. dQ/dt = -kQ

b. Q = 9e^{-kt}

c. k = 0.178

d. Q = 1.063 mg

Step-by-step explanation:

a) Write a differential equation for the quantity Q of hydrocodone bitartrate in the body at time t, in hours, since the drug was fully absorbed.

Let Q be the quantity of drug left in the body.

Since the rate of decrease of the quantity of drug -dQ/dt is directly proportional to the quantity of drug left, Q then

-dQ/dt ∝ Q

-dQ/dt = kQ

dQ/dt = -kQ

This is the required differential equation.

b) Solve your differential equation, assuming that at the patient has just absorbed the full 9 mg dose of the drug.

with t = 0, Q(0) = 9 mg

dQ/dt = -kQ

separating the variables, we have

dQ/Q = -kdt

Integrating we have

∫dQ/Q = ∫-kdt

㏑Q = -kt + c

Q = e^{-kt + c}\\Q = e^{-kt}e^{c}\\Q = Ae^{-kt}               (A = e^{c})

when t = 0, Q = 9

Q = Ae^{-kt}               \\9 = Ae^{-k0}\\9 = Ae^{0}\\9 = A\\A = 9

So, Q = 9e^{-kt}

c) Use the half-life to find the constant of proportionality k.

At half-life, Q = 9/2 = 4.5 mg and t = 3.9 hours

So,

Q = 9e^{-kt}\\4.5 = 9e^{-kX3.9}\\\frac{4.5}{9} = e^{-kX3.9}\\\frac{1}{2} = e^{-3.9k}\\

taking natural logarithm of both sides, we have

ln\frac{1}{2} = ln(e^{-3.9k})\\\\-ln2 = -3.9k\\k = -ln2/-3,9 \\k = -0.693/-3.9\\k = 0.178

d) How much of the 9 mg dose is still in the body after 12 hours?

Since k = 0.178,

Q = 9e^{-0.178t}

when t = 12 hours,

Q = 9e^{-0.178t}\\Q = 9e^{-0.178X12}\\Q = 9e^{-2.136}\\Q = 9 X 0.1181\\Q = 1.063 mg

7 0
3 years ago
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