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storchak [24]
3 years ago
10

Michael split a rope that was 28 inches long into 5 equal parts. Brenden split a rope that was 30 inches long into 6 equal parts

. Which boy's rope was cut into longer pieces?
Mathematics
1 answer:
QveST [7]3 years ago
7 0

Answer:

  Michael

Step-by-step explanation:

The pieces of Michael's rope were (28 in)/5 = 5.6 in long.

The pieces of Brenden's rope were (30 in)/6 = 5.0 in long.

Michael's rope was cut into longer pieces.

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Find the center and the radius of the circle with the equation:<br> x^2-4x+y^2
zysi [14]

Answer:

Step-by-step explanation:

8 0
3 years ago
You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which
stellarik [79]
To model this situation, we are going to use the compound interest formula: A=P(1+ \frac{r}{n} )^{nt}
where
A is the final amount after t years 
P is the initial deposit 
r is the interest rate in decimal form 
n is the number of times the interest is compounded per year
t is the time in years 

For account A: 
We know for our problem that P=2000 and r= \frac{2.25}{100} =0.0225. Since the interest is compounded monthly, it is compounded 12 times per year; therefore, n=12. Lets replace those values in our formula:
A=2000(1+ \frac{0.0225}{12} )^{12t}

For account B:
P=2000, r= \frac{3}{100} =0.03, n=12. Lest replace those values in our formula:
A=2000(1+ \frac{0.03}{12} )^{12t}

Since we want to find the time, t, <span>when  the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
</span>2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000

Now that we have our equation, we just need to solve for t:
2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000
(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}
(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5}&#10;{2}
ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})
12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})
t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})
t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}
17.47

We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>
5 0
3 years ago
An advertising company designs a campaign to introduce a new product to a metropolitan area of population 3 Million people. Let
Advocard [28]

Answer:

P(t)=3,000,000-3,000,000e^{0.0138t}

Step-by-step explanation:

Since P(t) increases at a rate proportional to the number of people still unaware of the product, we have

P'(t)=K(3,000,000-P(t))

Since no one was aware of the product at the beginning of the campaign and 50% of the people were aware of the product after 50 days of advertising

<em>P(0) = 0 and P(50) = 1,500,000 </em>

We have and ordinary differential equation of first order that we can write

P'(t)+KP(t)= 3,000,000K

The <em>integrating factor </em>is

e^{Kt}

Multiplying both sides of the equation by the integrating factor

e^{Kt}P'(t)+e^{Kt}KP(t)= e^{Kt}3,000,000*K

Hence

(e^{Kt}P(t))'=3,000,000Ke^{Kt}

Integrating both sides

e^{Kt}P(t)=3,000,000K \int e^{Kt}dt +C

e^{Kt}P(t)=3,000,000K(\frac{e^{Kt}}{K})+C

P(t)=3,000,000+Ce^{-Kt}

But P(0) = 0, so C = -3,000,000

and P(50) = 1,500,000

so

e^{-50K}=\frac{1}{2}\Rightarrow K=-\frac{log(0.5)}{50}=0.0138

And the equation that models the number of people (in millions) who become aware of the product by time t is

P(t)=3,000,000-3,000,000e^{0.0138t}

5 0
3 years ago
Which number line represents the solution set for the inequality –4(x + 3) ≤ –2 – 2x?
Yuliya22 [10]
<span>–4(x + 3) ≤ –2 – 2x

>>.....-4x -12 </span>≤ -2 -2x

>>  -12 +2 ≤ +2x 

>> - 10 ≤ 2x

>> -5 ≤ x............>> x >= -5 

This answer is not represented in the pictures you attached

The line starts in x = -5 and goes up to infinity
3 0
3 years ago
Read 2 more answers
Simplify the radical expression √(5) + 6√(5)
Evgesh-ka [11]
Answer is  <span>C) 7√(5)
cause

</span><span>√(5) + 6√(5)
</span>=<span>√5 (1+6)
</span>= 7<span>√5</span>
8 0
3 years ago
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