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3 years ago

ASAP PLS HELP ME WITH THIS PLZZZ

Mathematics

2 answers:

soldi70 [24.7K]3 years ago

8 0

Answer:

Step-by-step explanation:

6•11 for rectangle and 1/2 6•6 for triangle add and you get 66+18=84

sasho [114]3 years ago

8 0

Answer:

Step-by-step explanation:

find the area of the triangle first

= 1/2 * 6 * 6

=18 in^2

rectangle

= L * W

= 6*11

= 66 in^2

therefore, 66+18

=84

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If the measure of a triangle are in the ratio 2:5:5. What is the measure of the angle

Sphinxa [80]

Your answer is 75°, 75°, and 30°
Because you know that angles in a triangle equal 180°, you can find what 1 in terms of the ratio is by adding 2, 5, and 5, and then dividing 180 by that value. 2 + 5 + 5 = 12, so 180 ÷ 12 = 15.
Now that you know that 1 in terms of the ratio is 15, you can multiply 15 by the ratio values, so
15 × 2 = 30°
15 × 5 = 75°
15 × 5 = 75°

I hope this helps!

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4 years ago

Which is the completely factored from of xy^3-x^3y

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Answer:

$xy(y+x)(y-x)$

Step-by-step explanation:

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So the expression becomes:

$xy(y^2-x^2)\\xy(y+x)(y-x)$

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3 years ago

If you get a 3.1% raise and make 8 dollars an hour now, how much were you getting paid before

Nimfa-mama [501]

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Step-by-step explanation:

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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:

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