All categories

3 years ago

13.5 rounded to the nearest tenth

Mathematics

2 answers:

romanna [79]3 years ago

5 0

It’s already rounded to the nearest tenth

CaHeK987 [17]3 years ago

3 0

Answer:

iis just 13.5

Step-by-step explanation:

Please Mark Brainliest if i get the answer correct

You might be interested in

Write a two-step equation that involves multiplication and subtraction, includes a negative coefficient, and had a solution of x

Contact [7]

9x - 21 = -42
This should be the type of equation that you're looking 4

7 0

4 years ago

you are purchasing carpeting for an office building the space to be carpeted is 30 feet by 50 Company A charges a 2.99 per squar

vladimir2022 [97]

A:1500 x 2.99 +200= $4685
1500 ft^2=166.67 yard^2
B: 166.67 x 19.99+500= $3831.7333
= $3831.75
Company B is the better deal

3 0

4 years ago

Read 2 more answers

Use the quadratic formula to find the solutions to the quadratic equation below 2x2 - 3x + 6 = 0

Katarina [22]

Answer:

$x=\frac{3}{4} +i\frac{\sqrt{39}}{4}, x=\frac{3}{4} -i\frac{\sqrt{39}}{4}\\$

x=\frac{3}{4} ±i\frac{\sqrt{39}}{4}\\

Step-by-step explanation:

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

What is the area of the composite figure

oksano4ka [1.4K]

Answer:

Step-by-step explanation:

To find the area of a composite figure, separate it into the regular parts which make the irregular shape. This shape is composed of a semi-circle and a rectangle. Find the area by finding the area of each shape.

Semi-circle:

The semi-circle has a diameter of 2 + 4 + 2 = 8. The area this figure uses the radius which is half the diameter. The radius is 4. To find the area substitute r = 4 into $\pi r^2 = \pi 4^2 = 16\pi$ . However the semi-circle has a smaller circle cut out of it with radius 2. The area of the smaller circle is $\pir^2 = \pi 2^2=4\pi$ . The semi circle in the shape is the areas subtracted which equals 12π.

Rectangle:

The area of the rectangle is found using A = b*h = 2*5 = 10.

The total area is 12π + 10 meters squared.

5 0

3 years ago