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Fittoniya [83]
2 years ago
12

Dustin and Janelle sold packs of trading cards to earn extra money before summer break. They sold each box for $51 which include

d 6 packs of cards. Dustin sold n packs of trading cards, and Janelle sold 4 fewer packs than Dustin. They collected a combined total of $442. The following equation represents the given situation. 442 - How many packs of trading cards did Dustin sell? Dustin sold packs of trading cards.​
Mathematics
1 answer:
Grace [21]2 years ago
3 0

Answer:

Dustin sold 28 packs

Step-by-step explanation:

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Please help I’ve been stuck on it forever. :(
slava [35]

Answer:

$31.37 is the sale price

Step-by-step explanation:

All you need to do to find the discount price (sale price) is multiply the price of the item by the decimal form of the percentage.

1. Store A. 40% off the original price of $159.99

(159.99 * .40) = 63.996

63.996 is the discount price we are looking for

2. Store B. 25% off the original price of $130.50

(130.50 * .25) = 32.625

32.625 is the discount price we are looking for

3. Now to find their difference you just subtract the two values.

($63.996 - $32.625) = $31.371

The discount price (sale price) is $31.371

6 0
3 years ago
A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.
Usimov [2.4K]

Answer:

a) \frac{ds}{dt}= v(t) = 3t^2 -18t +15

b) v(t=3) = 3(3)^2 -18(3) +15=-12

c) t =1s, t=5s

d)  [0,1) \cup (5,\infty)

e) D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

f) a(t) = \frac{dv}{dt}= 6t -18

g) The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

Step-by-step explanation:

For this case we have the following function given:

f(t) = s = t^3 -9t^2 +15 t

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

\frac{ds}{dt}= v(t) = 3t^2 -18t +15

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

v(t=3) = 3(3)^2 -18(3) +15=-12

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

3t^2 -18 t +15 =0

We can divide both sides of the equation by 3 and we got:

t^2 -6t +5=0

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

(t-5)*(t-1) =0

And for this case we got t =1s, t=5s

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

t^2 -6t +5 >0

(t-5) *(t-1)>0

So then the intervals positive are [0,1) \cup (5,\infty)

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6

And if we replace we got:

D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

a(t) = \frac{dv}{dt}= 6t -18

Part g and h

The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

5 0
3 years ago
I have a question about basic algerba so I need some helps and its a word problem sooo yeah please help I'm giving out some big
Annette [7]

Answer:

2. x+0.3x

Step-by-step explanation:

hope this helps

4 0
3 years ago
Read 2 more answers
What is the ratio in simplest form?<br> 12 to 28
Oduvanchick [21]

Answer:

12:28 = 3:7

Step-by-step explanation:

Try to reduce the ratio further with the greatest common factor (GCF).

The GCF of 12 and 28 is 4

Divide both terms by the GCF, 4:

12 ÷ 4 = 3

28 ÷ 4 = 7

The ratio 12 : 28 can be reduced to lowest terms by dividing both terms by the GCF = 4 :

12 : 28 = 3 : 7

Therefore:

12 : 28 = 3 : 7

7 0
3 years ago
Which expression shows 2/3 + 4/5 written using a common denominator
lbvjy [14]

Answer:

10/15 + 12/15

Step-by-step explanation:

The LCD would be 15 and whatever you do to the top you do to the bottom

8 0
2 years ago
Read 2 more answers
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