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JulsSmile [24]
3 years ago
7

If the width of the rectangle is three times the lengthAnd a perimeter is 72 feet What are the length and width of the rectang

le
Mathematics
1 answer:
ira [324]3 years ago
4 0

Answer:

Width = 27

Length = 9

*You might of switched width and length around please check*

Step-by-step explanation:

Width = 3x

Length = x

Perimeter Formula

2(3x+x) = 72

2(4x) = 72

8x = 72

x = 9

Since X was found, plug it in to find the width and length

3(9) = 27

x = 9

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A Child’s Growth and Prosperity
MArishka [77]

Answer:

a) 0.75 inch/month

b) y-20 = 0.75(x-4)

c) 17 inches

d) 44

a) 0.8 inch/month

b) yes

c) y = 0.8x + 18.8

d) 18.8 inches

Step-by-step explanation:

a) (21.5-20)/(6-4) = 1.5/2

= 0.75 inch/month

If this is correct, 2 months after the 6th month (month 8), it should be

21.5 + 2(0.75) = 23

Which it is

b) y-y1 = m(x-x1)

Where y is the length, x is the months, m is the slope and (x1,y1) is a known point

At x = 4, y = 20

y-20 = 0.75(x-4)

Or

y = 0.75x + c

20 = 0.75(4) + c

c = 20 - 3 = 17

y = 0.75x + 17

c) at birth, 4 months before month 4

So, 20 - 0.75(4) = 17

Which is also the y-intercept

d) y = 0.75(36) + 17

y = 27 + 4 = 44 inches

Generally, at the age of 3 years (36 month) the length is between 35 and 40 inches

a) slope = (30-22)/(14-4)

= 0.8 inch/month

b) 0.8 > 0.75, so yes

c) When x = 4, y = 22

y-22 = 0.8(x - 4)

y - 22 = 0.8x - 3.2

y = 0.8x + 18.8

y-intercept is 18.8

d) at birth, x = 0.

Length was 18.8 inches

Step-by-step explanation:

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The answer is 1.1 x10^7
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30,77,55,71,19,40,10,5,7,6
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Gina puts $ 4500 into an account earning 7.5% interest compounded continuously. How long will it take for the amount in the acco
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~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5150\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years \end{cases}

5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}

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Answer:

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Step-by-step explanation:

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