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Natali5045456 [20]
3 years ago
6

The price of milk has been steadily increasing 5% per year. If the cost of a gallon is now $3.89: What will it cost in 10 years?

What did it cost 5 years ago?
Mathematics
1 answer:
saw5 [17]3 years ago
3 0
This is the concept of exponential growth; The rate of growth of the price milk is 0.05, current price=$ 3.89;
this information can be modeled by the exponential growth equation given by:
y=ae^(nr)
where;
a=initial value
r=rate of growth;
n=number of years
thus the function modeling our information will be:
y=3.89e^0.05n

the price after 10 years will be:
y=3.89e^(0.05*10)
y=3.89*1.649
y=6.41

the price after 10 years will be $ 6.41

The price 5 years ago will be:
y=3.89*e^(-5*0.05)
y=3.89 e^(-0.25)
y=3.03
thus we conclude that the price 5 years ago was $ 3.03
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374 × 510<br><br> What are the partial products Lucas will need to solve the problem?
BabaBlast [244]
The partial products that Lucas would need to solve this problem would be 3,740 and 187,000 if you are multiplying 374 on the top column and 510 on the bottom column. Or, the partial products Lucas would need is 2,040 and 35,700 and 153,000. Both partial products combination added together would be 190,740.
6 0
3 years ago
Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

3 0
3 years ago
HELP PLEASE ASAP
Irina18 [472]
D is your answer if you times 8 and 2 together to get your A& B
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I need a answer asap
Murrr4er [49]

Answer:

As the weight increases, the price increases.

6 0
3 years ago
Read 2 more answers
Guyssssssss help me 10 points!!
zlopas [31]

Answer:

19

Step-by-step explanation:

the side is same as 57

what times 3 is 57?

57/3 is 19

AB is 19

pls vote me as brainliest!!!<3

8 0
3 years ago
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