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3 years ago

İs this photo fake know??Come to comments

Mathematics

2 answers:

scZoUnD [109]3 years ago

7 0

Answer:

Looks like something you would find in a 6th graders photo album

Step-by-step explanation:

yeet

vova2212 [387]3 years ago

6 0

Answer:

CRINGE

Step-by-step explanation:

Ok ngl BIG BIG CRINGE

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Given the following vector fields and oriented curves C, evaluate integral F * T dsF = <-y, x> on the semicircle r(t) = &l

luda_lava [24]

Answer:

Answer is 16 $\pi$ .

Step-by-step explanation:

4 0

3 years ago

Please answer this question correctly I need it today please show work

oksano4ka [1.4K]

Answer:

1: C(n) = 2.50 + 16n

2: $66.50

Step-by-step explanation:

Part 1

Each ticket costs $16 per person. If tickets for n persons were purchased, the total cost would be 16n.

There is also a one-time service fee of $2.50 that must be paid. Thus, for n tickets the total cost is

C(n) = 2.50 + 16n

Part 2

For n = 4, the expression evaluates to

C(4) = 2.50 + 16 (4) = $66.50

7 0

3 years ago

How can you rewrite the expression (8-5i)^2 in the form a+bi?

otez555 [7]

(a-b)^2 = a^2-2ab+b^2

(8-5i)^2 = 8^2-2(8)(5i)+(5i)^2

= 64-80i+25i^2

i^2=-1

So

= 64-80i+25(-1)

=64-25-80i

= <em><u>39 - 80i</u></em>

which is your answer :)

8 0

3 years ago

Read 2 more answers

PLEASE HURRY!! TIMED!! WILL GIVE BRAINLIEST!!!

tensa zangetsu [6.8K]

Answer:

2/3 +9.26 = 0.6666+9.26=rational

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