Ask question

Ask question

All categories

3 years ago

12

Solve for X

s="latex-formula">

2 answers:

Musya8 [376]3 years ago

5 0

Answer:

x = 5

Step-by-step explanation:

To solve, first subtract 5 from each side:

5 + 7x = 40

7x = 35

Then divide each side by 7:

x = 5

just olya [345]3 years ago

4 0

Answer:

5

Step-by-step explanation:

You might be interested in

<img src="https://tex.z-dn.net/?f=Maria%20earns%20%24603.75%20for%2035%20hours%20of%20work.%20What%20is%20her%20rate%20of%20pay%

Solnce55 [7]

Answer:$17.25 is her rate per hour

Step-by-step explanation:

Do 603.75/35

Which $17.25

4 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

The parabola y=ax^2+bx-10 passes through the points (4.5,8) and (-2.5, 15). Determine the values of a and b

yuradex [85]

$y=ax^2+bx-10 \\ \\
(4.5,8) \\
8=a \times 4.5^2 + b \times 4.5-10 \\
8=20.25a+4.5b-10 \ \ \ |+10 \\
18=20.25a+4.5b \\ \\
(-2.5,15) \\
15=a \times (-2.5)^2+b \times (-2.5)-10 \\
15=6.25a-2.5b-10 \ \ \ |+10 \\
25=6.25a-2.5b$

The system of equations:
$18=20.25a+4.5b \ \ \ |\times 5 \\
25=6.25a-2.5b \ \ \ |\times 9 \\ \\
90=101.25a+22.5b \\
\underline{225=56.25a-22.5b} \\
90+225=101.25a+56.25a \\
315=157.5a \ \ \ |\div 157.5 \\
a=2 \\ \\
25=6.25a-2.5b \\
25=6.25 \times 2-2.5b \\
25=12.5 -2.5b \ \ \ |-12.5 \\
12.5=-2.5b \ \ \ |\div (-2.5) \\
b=-5 \\ \\
\boxed{a=2} \\ \boxed{b=-5} \\ \boxed{y=2x^2-5x-10}$

6 0

3 years ago

These two scalene triangles are similar with a scale of 7:4 what is true about these figures

erica [24]

we know that

$scale\ factor =\frac{7}{4}$

if triangle GHJ and triangle MKN are similar

then

$\frac{GH}{MK} =\frac{HJ}{KN}= \frac{GJ}{MN} = \frac{7}{4}$

m∠g°=m∠m°

m∠h°=m∠k°

m∠j°=m∠n°

therefore

<u>the answer is</u>

$\frac{GJ}{MN} = \frac{7}{4}$

6 0

3 years ago

Read 2 more answers

You have $437 in a savings account. After a deposit, the balance is $1087. What was the amount of the deposit?

Licemer1 [7]

Answer 650

Explanation: subtract 437 from 1087

6 0

4 years ago