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2 years ago

If you answer this I will mark you brainliest

Mathematics

1 answer:

Temka [501]2 years ago

8 0

Answer:

Here is the ans..hope it helps:)

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An engineer is planning a new water pipe installation. The circular pipe has a diameter of d=20\text{ cm}d=20 cmd, equals, 20, s

aivan3 [116]

Answer: The answer is 314.28 cm² (approx.).

Step-by-step explanation: Given that an engineer is going to install a new water pipe. The diameter of this circular pipe is, d = 20 cm.

We need to find the area 'A' of the circular cross-section of the pipe.

Given, diameter of the circular section is

$\textup{d}=20~\textup{cm}.$

So, the radius of the circular cross-section will be

$\textup{r}=\dfrac{\textup{d}}{2}=\dfrac{20}{2}=10~\textup{cm}.$

Therefore, cross-sectional area of the pipe is

$\textup{A}=\pi \textup{r}^2=\dfrac{22}{7}(10)^2=\dfrac{2200}{7}=314\dfrac{2}{7}=314.28~.~.~.~\textup{cm}^2.$

Thus, the answer is 314.28 cm² (approx.).

4 0

3 years ago

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

3 years ago

The difference between the squares of two consecutive numbers is 23. what are the two numbers?

Ksju [112]

(x+1)^2-x^2=23

x^2+2x+1-x^2=23

2x+1=23

2x=22

x=11

11^2 = 121

x+1 =12

12^2=144

144-121 = 23

4 0

3 years ago

Simplify the product of the binomial. (5t+4)^2

krek1111 [17]

The answer would be 25t^2 + 40t + 16

8 0

3 years ago

Read 2 more answers

Helen told her mom she got 85% on a recent math test. She remembered the test was out of 60. What was her actual mark? *

barxatty [35]

Answer: 85% out of how many questions.

Step-by-step explanation:

8 0

3 years ago