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adelina 88 [10]
3 years ago
5

Maggie leaves her home and travels 4 miles

Mathematics
1 answer:
DiKsa [7]3 years ago
4 0

Answer:11 miles

Step-by-step explanation:

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Angelina_Jolie [31]

Answer:

15.13cm

Step-by-step explanation:

From that above question:

We are told that:

Lateral surface area of a right circular cone : πr√r² + h²

We are give the following parameters:

Lateral surface area = 236.64cm²

Radius = 4.75cm

We are to find the height

Making h the subject of the formula

h = √[(A/r)² - πr²]/π

h = √[(236.64/4.75)²- π ×4.75²]/π

h = 15.12975 cm

Approximately to the nearest hundredth = 15.13cm

7 0
3 years ago
I buy a MP3 player for £40 in a sale where it was 20% off. What was the original price
natta225 [31]
20% off means you paid only 80% of the original price.
0.8x=40
x=40÷0.8=50
so the original price is 50
6 0
3 years ago
Read 2 more answers
10 points + brainless to first correct answer!<br> Total 18 points.
butalik [34]

Answer:

8

Step-by-step explanation:

hope it help toyou

love

6 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=a%20-%207%20%3D%2017" id="TexFormula1" title="a - 7 = 17" alt="a - 7 = 17" align="absmiddle" c
andriy [413]
A would equal 24 because 17+7 is 24
6 0
3 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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