4. A number cube with the number 1 through 6 is

|3k-2|=2|k+2| how do I do this when the 2 is on the other side??

grin007 [14]

|3k-2|=2|k+2l
Do Distributive Property and take 2 times by everything in the parenthesis. That will leave you with |3k-2|=2k+4. Then add like terms. 3k+2k=5k. Then -2+4=2. So you end up with 5k+2. I believe I'm right. If not please tell me.

3 0

3 years ago

Someone please help me with these too://

algol [13]

Answer:

1.

The slope is 2

The y-intercept is -10

equation: y = 2x - 10

2.

Slope: 2.5

y-intercept: 1

Equation: y = 2.4x + 1

Step-by-step explanation:

1.

x y

2 12

4 16

8 24

10 28

from 12 to 16 you add 4

from 2 to 4 you add 2

y/x = 4/2 = 2

The slope is 2

The y-intercept is -10

equation: y = 2x - 10

I got the y-intercept because I went from 12 to 0 which is -12, and then 2 - 12 and it's -10

__________________________________________________________

2.

x y

5 4

20 10

30 14

35 16

from 4 to 10, u add 6

from 5 to 20 you add 15

y/x = 15/6 = 2.5

Slope: 2.5

y-intercept: 1

Equation: y = 2.4x + 1

I got the y-intercept because I went from 4 to 0 which is -4, and 5 - 4 is 1.

Hope this helped

7 0

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

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

3 years ago

The point(5/13,Y) IS IN THE 4TH quadrant corresponds to angle on the unit circle

shutvik [7]

4th quadrant is positive x (cos) and negative y (sin)

Use the Pythagorean Theorem to calculate the value of y (sin).
x² + y² = c²
5² + y² = 13²
25 + y² = 169
y² = 144
y = 12

Since the y-value is negative in the 4th quadrant, then y = -12

6 0

4 years ago

Which point on the nuumber line represents the product (-4)(-2)(-1)

drek231 [11]

Answer:

-8

Step-by-step explanation:

(-4)(-2)(-1)

8(-1)

-8

6 0

3 years ago