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Mama L [17]
2 years ago
12

Pls answer 26 and show work

Mathematics
1 answer:
Phoenix [80]2 years ago
7 0

The transformed equation y = -(x - 2)^2 - 3 compared to the parent function involves  translating the parent function to the right by 2 units, reflecting the function across the y-axis and translating the function 3 units down

<h3>How to compare the function to its parent function?</h3>

The equation of the transformed function is given as:

y = -(x - 2)^2 - 3

While the equation of the parent function is given as

y = x^2

Start by translating the parent function to the right by 2 units.

This is represented as:

(x, y) = (x - 2, y)

So, we have:

y = (x - 2)^2

Next, reflect the above function across the y-axis

This is represented as:

(x, y) = (-x, y)

So, we have:

y = -(x - 2)^2

Lastly, translate the above function 3 units down

This is represented as:

(x, y) = (x, y - 3)

So, we have:

y = -(x - 2)^2 - 3

Hence, the transformed equation y = -(x - 2)^2 - 3 compared to the parent function involves  translating the parent function to the right by 2 units, reflecting the function across the y-axis and translating the function 3 units down

Read more about function transformation at:

brainly.com/question/8241886

#SPJ1

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Step-by-step explanation:

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<em>0</em>

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<em>3</em><em>x</em><em>=</em><em>0</em>

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<em>hope </em><em>it </em><em>helps</em>

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storchak [24]
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3 years ago
At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given
Nostrana [21]

Answer:

a. The initial amount was 10 mg.

b. The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.

c. The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.

d.  The time until only 1 mg of the drug remains in the body is 11.6 hours.

Step-by-step explanation:

You know that at time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by :

A=10*(0.82)^{t}

a. The initial quantity occurs when time t is the initial t, that is, t is equal to 0. Then:

A=10*(0.82)^{t}=10*(0.82)^{0}=10*1\\

A=10

<u><em>The initial amount was 10 mg.</em></u>

b. Considering that an exponential growth is determined by:

A=A0*(1-r)^{t}, where A is the amount after a certain number of a certain time, Ao is the initial amount, r is the rate and t is the time, so the percentage of the drug that leaves the body each hour is :

1-r=0.82

Solving:

1-r -1= 0.82 -1

-r= -0.18

r= 0.18

<u><em>The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.</em></u>

c. The amount of drug that remains in the body 6 hours after dosing is when t = 6:

A=10*(0.82)^{6}

Solving:

A= 3.04 mg

<u><em>The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.</em></u>

d. The time that passes until only 1 mg of the drug remains in the body is calculated taking into account that A = 1 mg:

1=10*(0.82)^{t}

Solving:

0.1=(0.82)^{t}

㏒ 0.1= t*㏒ 0.82

㏒ 0.1  ÷ ㏒ 0.82= t

11.6 hours= t

<u><em> The time until only 1 mg of the drug remains in the body is 11.6 hours.</em></u>

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