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Sergeu [11.5K]
3 years ago
12

Janesa has 24 coins in her pocket word exactly $1 she tells her younger brother Jackson that she can have the coins if you go di

rectly guess how many of each corner is she gives them a hint that are only nickels and pennies
Mathematics
2 answers:
denis23 [38]3 years ago
8 0

Answer:

19 nickels and 5 pennies

Step-by-step explanation:

Let x nickels and y pennies

100 = 5x + y

x + y = 24

y = 24 - x

100 = 5x + 24 - x

4x = 76

x = 19

y = 24 - x = 24 - 19 = 5

IgorLugansk [536]3 years ago
8 0

Answer:19 Nickles

Step-by-step explanation:

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BabaBlast [244]

Ans: 144

Step-by-step explanation:

24 pounds per foot.

120+24=144

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3 years ago
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snow_lady [41]

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Step-by-step explanation:

4/12 = 1/3

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PLEASE HELP ME PLEASE HELP ASAP PLEASE HELP PLEASE AND THANK YOU
Allushta [10]

Answer:

y =mx + c

7 = 5/2 (-2) + c

7 = - 5 + c

c = 7 + 5

c = 12

Therefore, y = 5/2x + 12

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7 0
3 years ago
The rate at which the temperature is dropping is 6t 5t2 degrees Celsius per hour t hours after sundown. How much had the tempera
Whitepunk [10]

Using a differential equation, it is found that the temperature dropped 72 degrees Celsius between sundown and 3 hours after sundown.

<h3>What is the differential equation that describes the temperature in t hours after sundown?</h3>

The rate at which the temperature is dropping is 6t + 5t^2 degrees Celsius per hour t hours after sundown, hence the <em>differential equation</em> is:

\frac{dT}{dt} = 6t + 5t^2

Applying <em>separation of variables</em>, we find the solution as follows:

\frac{dT}{dt} = 6t + 5t^2

dT = (6t + 5t^2) dt

\int dT = \int (6t + 5t^2) dt

T(t) = \frac{5t^3}{3} + 3t^2 + T(0)

In which T(0) is the temperature at sundown.

In 3 hours, the change will be of:

C = T(3) - T(0) = \frac{5t^3}{3} + 3t^2|_{t = 3}

Hence:

\frac{5(3)^3}{3} + 3(3)^2 = 45 + 27 = 72

The temperature dropped 72 degrees Celsius between sundown and 3 hours after sundown.

You can learn more about differential equations at brainly.com/question/24348029

5 0
2 years ago
find the spectral radius of A. Is this a convergent matrix? Justify your answer. Find the limit x=lim x^(k) of vector iteration
Natasha_Volkova [10]

Answer:

The solution to this question can be defined as follows:

Step-by-step explanation:

Please find the complete question in the attached file.

A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]

now for given values:

\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\

\to  (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to  (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} -  \frac{1}{4}] =0 \\\\\to  (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\

\to \lambda_1=\lambda_2 =\frac{1}{2}\\\\\to \lambda_3 = \frac{1}{4} \\\\\to A = max {|\lambda_1| , |\lambda_2|, |\lambda_3|}\\\\

       = max{\frac{1}{2}, \frac{1}{2}, \frac{1}{4}}\\\\= \frac{1}{2}\\\\(A) =\frac{1}{2}

In point b:

Its  

spectral radius is less than 1 hence matrix is convergent.

In point c:

\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) =   \left(\begin{array}{c}3&1&2\end{array}\right)  , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\  \to x^{(k+1)} =  \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right]  \\\\

after solving the value the answer is

:

\lim_{k \to \infty} x^k=o  = \left[\begin{array}{c}0&0&0\end{array}\right]

4 0
3 years ago
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