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TEA [102]
3 years ago
13

Answers A) 20,950.70 B) 28,458.50 C) 31,045.50 D) 34,650.75

Mathematics
1 answer:
IgorC [24]3 years ago
7 0
A) 20,950.70
b) 28,458.50
c) 31,045.50
d) 34,650.75
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1.5x - 1.9y = - 29 , x - 0.9 = 4.5
Katyanochek1 [597]
X-0.9 = 4.5
+0.9 +0.9
x = 5.4
( to get rid of the -0.9 beside the x, u have to do the opposite operation on both sides. if its -0.9, u have to add 0.9 on both sides)

not sure about the first one.
3 0
3 years ago
Are the two figures similar? D 6 15 9 c​
Viktor [21]

Answer:

These are similar. The scale factor is 1.5 going from ABC to DEF.

Step-by-step explanation:

What I did was divide the side lengths of DEF by the corresponding side on ABC.

12/8 = 1.5

9/6 = 1.5

15/10 = 1.5

Since the scale factor is the same on all three sides, these two figures are similar.

5 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

<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>

4 0
3 years ago
6 &lt; 3x + 9 ≤ 18<br><br><br> please help me I will mark brainiest answer thanks
Olin [163]

Answer:

Step-by-step explanation:

6< 3x + 9

-3x < 9-6

-3x < 3 .( -1 )                                            

3x > 3                          

x > 3/3

x > 3

3x + 9 ≤ 18

3x ≤ 18-9

3x ≤ 9

x ≤ 9/3

x ≤  3

S = {XeIR/ 3 > x  ≤ 3 }

Not sure if the solution looks like this

3 0
3 years ago
Read 2 more answers
Simplified <br> 2(x + 8) + 3(x -9)
slava [35]

Answer:

5x - 11

Step-by-step explanation:

Step 1: Write out expression

2(x + 8) + 3(x - 9)

Step 2: Distribute

2x + 16 + 3x - 27

Step 3: Combine like terms (x)

5x + 16 - 27

Step 4: Combine like terms (constants)

5x - 11

5 0
3 years ago
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