All categories

4 years ago

Darius wrote 17- 4=225 what is his error

Mathematics

2 answers:

podryga [215]4 years ago

8 0

Well the answer would be 13, because 13 plus 4 is 17.

timofeeve [1]4 years ago

7 0

17-4 is actually 13 so the percentage error is $\frac{|actual-error|}{actual} * 100$

You might be interested in

Use calculator to evaluate each expression. (PICTURE) round to nearest hundredth.

adoni [48]

Answer:

cos 85 = .09

tan 18 =.32

sin 41=.66

Step-by-step explanation:

cos 85 = .087155743 = .09

tan 18 =.324919696=.32

sin 41=.656059029=.66

4 0

3 years ago

2x + y = -5 how do I solve this

ozzi

Answer:

if you are solving for x the answer is -5/2 and for y answer is 5

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Consider a water pipe that branches into two smaller pipes. If the flow of water is 10 L/min in the main pipe and 4 L/min in one

masha68 [24]

The water flows 6 liters per minute in the other branch of pipe.

Step-by-step explanation:

Given,

Flow of water in main pipe = 10L/min

Flow of water in one branch = 4L/min

Flow of water in other branch = x

Flow of water in main pipe = Flow in one branch + Flow in other branch

$10=4+x\\10-4=x\\6=x\\x=6$

The water flows 6 liters per minute in the other branch of pipe.

4 0

4 years ago

Find the area of a regular octagon with an apothem of 8.45 cm and a side length of 7 cm. (round to the nearest whole number) A)

Alexeev081 [22]

Answer:

237

Step-by-step explanation:

5 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