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baherus [9]
3 years ago
12

1. SUNDAES Carmine bought 5 ice cream sundaes for his friends. If each sundae costs $4.95, how much did he

Mathematics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

24.75

Step-by-step explanation:

you want to multiply 4.95×5 and you will get the answer $24.75.

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21) Find x. Show your work.
shepuryov [24]

Answer:

x = 21

Step-by-step explanation:

Givens:

Every triangle's interior angles add up to 180°

The triangle has a right angle.

The triangle is isosceles (two equal sides and two equal angles)

Let's make an equation with the information we have already.

2 × (2x + 3) + 90 = 180

4x + 6 + 90         = 180

4x + 96                = 180

4x                         = 84

x                           = 21

4 0
3 years ago
Which shape has 2 parallel sides ready
Irina-Kira [14]

Answer:

You have two sets of parallel sides. A square has two sets of parallel sides, and it has the extra condition that all of the angles are right angles. So a square is definitely going to be a rhombus. Now, all rhombuses have four sides.

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
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
Help please i need to know this
Harlamova29_29 [7]

Answer:

As shown in the attachment,

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Find the cube of a+3​
Sedbober [7]

Answer:

(a+3)^3

a^3 + b^3 + 3ab [identity : (a+b)^3 )]

a^3+ 3^3 + 3×a×3

a^3 +27+9a

answer : a^3+27+9a

hope it helps you

mrk me braniliest plz

6 0
3 years ago
Read 2 more answers
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