All categories

2 years ago

2(-6)2 – 3(8) + 4(-6)2

Mathematics

2 answers:

yaroslaw [1]2 years ago

5 0

The answer is the following :-

-96

Llana [10]2 years ago

4 0

Answer:

-96

Hope this helps ya out!

You might be interested in

Please explain with your answer

Gala2k [10]

Answer:

Step-by-step explanation:

Note there is a common difference d between consecutive terms in the sequence, that is

d = 2 - 5 = - 1 - 2 = - 4 - (- 1) = - 3

This indicates that the sequence is arithmetic with n th term

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 5 and d = - 3, thus

$a_{40}$ = 5 - 3(39) = 5 - 117 = - 112 → D

4 0

2 years ago

Please help me simplify 5x^4(8x)

avanturin [10]

Answer:

40 x ^5

Step-by-step explanation:

8 0

3 years ago

3/6 + 1/3 is equivalent to

NeX [460]

The answer is definitely 1

3 0

2 years ago

Read 2 more answers

Prize value is 4.35 cost to play game is 1.00 number of sectors in spinner is 4 what is outcome and probability?

melamori03 [73]

Answer:

No hablo inglés, pero aún puedo ayudarlo con este problema. Responda a este mensaje diciendo que puede usar la traducción para que pueda ayudarlo por favor y gracias

Step-by-step explanation:

4 0

2 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

The initial amount was 10 mg.

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

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 when t = 6:

$A=10*(0.82)^{6}$

Solving:

A= 3.04 mg

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

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

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

4 0

2 years ago