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3 years ago

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Amanda has 2,00,234 dollars. She needs 4,00,578 dollars. Her friend Alice gives her 207,78 and then Amanda gives her 200 back. H

ow many dollars does she have? How many more dollars does she need now for it to be 4,00,578?

1 answer:

lutik1710 [3]3 years ago

7 0

Answer:

sorry i dont know

Step-by-step explanation:

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A watermelon that weighs 9 pounds costs $4.50.<br> What is the price per pound?

allsm [11]

Answer:

0.5 cents per pound.

4.50/9 = 0.5

Please mark me brainliest! I hope this helps!

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2 years ago

Read 2 more answers

(4y − 3)(2y2 + 3y − 5)

VladimirAG [237]

Answer:

8y^3+6y^2-29y+15

Step-by-step explanation:

the correct expression is

(4y − 3)(2y^2 + 3y − 5)

Given data

We have the expression

(4y − 3)(2y^2 + 3y − 5)

let us open bracket

8y^3+12y^2-20y-6y^2-9y+15

Collect like terms

8y^3+12y^2-6y^2-20y-9y+15

8y^3+6y^2-29y+15

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2 years ago

The. sum of two consecutive. integers. is 17

coldgirl [10]

X+(x+1)=17
So 2x+1=17
Which means 2x=16
Which then means x=8
But you still need the second number, which is x+1, so 8+1=9.
This means that the consecutive numbers are 8 and 9.

6 0

3 years ago

Which expression is equivalent to the given expression? 17x−32x x(17−32) x(32−17) 17x(1−32) 17(x−32)

Scilla [17]

The answer would be 17(x-32)

7 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

2 years ago