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Naddik [55]
4 years ago
7

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
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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
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