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2 years ago

Do y'all know vectors-? AHH

Mathematics

2 answers:

Sophie [7]2 years ago

3 0

Answer:

yea kind of but not in detail

grigory [225]2 years ago

3 0

You mean math vectors?

You might be interested in

In a classroom,

Gelneren [198K]

Answer:

3 students in blue and 12 in white

Step-by-step explanation:

set 1/6 and 2/3 of it and you need to find that of 18

so you take 2/3*18 and get 12 students

and 1/6*18 and get 3 students

7 0

2 years ago

Which statement is true about ∠1 and ∠2?

gavmur [86]

Answer: Most likely A

Step-by-step explanation:

the reason I say its A is because of the face that congruent means in line not and straight. Though in your answer it shows that they are not alined so that leaves us with complemantary and the 2 complementary answer choices are

A & C but C is wrong because it say alternate which also means the that they would be 2 different angels so yeah I'm gonna go for A.

Good luck :)

5 0

2 years ago

triangle abc has vertices at a(-2 3),B(0,3),c(-1,-1). Find the coordinates of the image after a reflection over the x-axis.

Crazy boy [7]

The new coordinates of the triangle after reflection from x-axis would be A(-2, -3) , B(0, -3) and C( -1, 1 )

<u>Explanation:</u>

The vertices of the triangle are A(-2 3), B(0,3), C(-1,-1)

The reflection is over the x-axis.

When the reflection is along the x-axis, the coordinates of x does not change. However, the coordinates of y does change.

The negative value of y becomes positive and the positive value of y becomes negative.

Similarly,

A(-2, 3) would become A(-2, -3)

B(0, 3) would become B(0, -3)

C(-1, -1) would become C( -1, 1 )

Therefore, the new coordinates of the triangle after reflection from x-axis would be A(-2, -3) , B(0, -3) and C( -1, 1 )

6 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

Simplify.<br><br> 4 x (8 + 5) + 9 can someone help me out plz

Xelga [282]

Step-by-step explanation:

4x(8+5)+9

=52x+9

Description:

The first step is to simplify the equation provided. So in this case the equation is 4 x (8 + 5) + 9. After that you will get your answer as =52x+9. So in this case 4 x (8 + 5) + 9 = =52x+9.

Answer: =61

Hope this helps.

8 0

2 years ago

Read 2 more answers