Solve the Proportion. 3/4=x/18

Suppose 4 mg of a drug is injected into a person's bloodstream. As the drug is metabolized, the quantity diminishes at the conti

Brut [27]

Let Q(t) = the mass (mg) remaining after t hours.
We are told that Q diminishes by 3% every hour.

When t = 0, Q = 4 mg
When t = 1, Q = 4*0.97 mg
When t = 2, Q = 4*(0.97)² mg

By induction,
$Q(t) = 4 (0.97)^{t} \, mg$

Q'(t) = 0.97Q
Therefore the rate of decrease is 3% per hour.

The person receives an additional dosage when Q falls to 0.50 mg. This happens when
$4(0.97)^{t}=0.5 \\0.97^{t} = 0.125 \\ t \,ln(0.97) = ln(0.125) \\ t = \frac{ln(0.125)}{ln(0.97)} =68.27 \, hrs$

After the second injection, the mass is now 4.5 mg. Therefore
$Q(t)=4.5(0.97)^{t}$

When the mass again reaches 0.50 mg, then
$4.5(0.97)^{t}=0.5 \\ 0.97^{t}=0.1111 \\ t \, ln(0.97) = ln(0.1111) \\ t = 72.14 \, hrs$

Answers:
(a) $Q(t) = 4(0.97)^{t}$
(b) 3% per hour
(c) 68.3 hours
(d) 72.1 hours

3 0

3 years ago

Can someone explain Part B to me please? I'm trying to practice for SBAC but I don't get this.

Finger [1]

On part A you put 1/2 being greater, so part B would be the opposite which would be 1/8

3 0

2 years ago

Luke’s basketball team went to an amusement park at the end of the season. The cost of the admission for 5 coaches and 12 player

Finger [1]

Answer:

The admission cost for each player was 22.50.

Step-by-step explanation:

Given:

Luke’s basketball team went to an amusement park at the end of the season.

The cost of the admission for 5 coaches and 12 players was 407.50.

The admission cost for each coach was 27.50.

Now, to find the admission cost for each player.

Total admission cost = 407.50.

Number of coaches = 5.

Admission cost for each = 27.50.

So, the cost of admission of all coaches:

$27.50\times 5$

$=137.50.$

Then, we deduct the cost of admission of all coaches from the total admission cost:

$407.50-137.50$

$=270.$

Remaining cost = 270.

Number of player = 12.

Now, to get the admission cost for each player we divide the remaining cost by number of players:

$270\div 12$

$=22.50.$

Therefore, the admission cost for each player was 22.50.

7 0

2 years ago

What is the value of 4*(-1+2-3+4-3+6-7+....+100) Show your work. [Pls help I am really drowning in work. 50 points + brainliest

KengaRu [80]

Answer:

200

Step-by-step explanation:

We have:

$4(-1+2-3+4-5+6-7...+100)$

We can rearrange the numbers to obtain:

$4((-1-3-5-7-...-99)+(2+4+6...+100))$

From the left, we can factor out a negative. So:

$4(-(1+3+5+7+...+99)+(2+4+6...+100))$

In other words, we want to find the sum of all the odd numbers from 1 to 99.

And the sum of all the even numbers from 2 to 100.

Let's do each one individually:

Odd Terms:

We have:

$(1+3+5+7+...+99)$

We can use the arithmetic series formula, where:

$S=\frac{k}{2}(a+x_k)$

Where k is the number of terms, a is the first term, and x_k is the last term.

Since it's all the odd numbers between 1 and 99, there are 50 terms.

Our first term is 1 and our last term is 99. So, the sum of all the odd terms are:

$S=\frac{50}{2}(1+99})$

Divide the fraction. Add within the parentheses:

$S=25(100)$

Multiply:

$S=2500$

So, the sum of all the odd terms is 2500.

Even Terms:

We have:

$(2+4+6+...+100)$

Again, we can use the above formula.

Our first term is 2, last term is 100. And since it's from 2-100, we have 50 even terms. So:

$S=\frac{50}{2}(2+100)$

Divide and add:

$S=25(102)$

Multiply:

$S=2550$

We originally had:

$4(-(1+3+5+7+...+99)+(2+4+6...+100))$

Substitute them for their respective sums:

$4(-(2500)+2550)$

Multiply:

$4(-2500+2550)$

Add:

$=4(50)$

Multiply:

$=200$

So, the sum of our sequence is 200.

And we're done!

Note: I just found a way easier way to do this. We have:

$4\cdot(-1+2-3+4-5+6-7+...+100)$

Let's group every two terms together. So:

$=4((-1+2)+(-3+4)+(-5+6)...+(-99+100))$

We can see that they each sum to 1:

$=4((1)+(1)+(1)+...+(1))$

Since there are 100 terms, we will have 50 pairs, so 50 times 1. So:

$=4(50)$

Multiply:

$=200$

Pick which one you want to use! I will suggest this one though...

Edit: Typo

8 0

3 years ago

I have a total of 29 coins-they are quarters and dimes. Together they amount to $5.60. How many quarters do I have? How many dim

Anna007 [38]

450/25= 18

where 450 is $4.50 and 25 is one quarter. there will be 18 quarters.

560-450=110

110/10=11

where 110 is $1.10 and 10 is one dime. there will be 11 dimes.

11+18= 29

where 11 is total number of dimes and 18 is total number of quarters. and 29 is total number of coins

6 0

2 years ago

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