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pogonyaev
3 years ago
12

Find x and y please

Mathematics
2 answers:
Alexandra [31]3 years ago
6 0

Answer:

Answer:

  • x → 4.24
  • y → 4.24

Step-by-step explanation:

» x and y can only be got through using trigonometric ratios:

✍ <u> </u><u>For</u><u> </u><u>x</u><u> </u>:

{ \tt{ \sin( \theta)  =  \frac{opposite}{hypotenuse} }} \\  \\ { \tt{ \sin(45 \degree)  =  \frac{x}{6} }} \\  \\ { \tt{x = 6 \times  \sin(45 \degree) }} \\ { \tt{x = 4.24}}

✍ <u> </u><u>F</u><u>o</u><u>r</u><u> </u><u>y</u><u> </u>:

{ \tt{ \cos( \theta)  =  \frac{adjacent}{hypotenuse} }} \\  \\ { \tt{ \cos( 45  \degree)  =  \frac{y}{6} }} \\  \\ { \tt{y = 6 \times  \cos(45 \degree) }} \\ { \tt{y = 4.24}}

natulia [17]3 years ago
5 0

Answer:

see below

Step-by-step explanation:

this is a 45 90 45-45-90 right triangle. the side is x and the hypotenuse is x*(sqrt(2)). the hypotenuse is 6, so the side is 3(sqrt(2)).

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8 0
3 years ago
Which fraction has a value that's equal to 7/8
S_A_V [24]
There's a bunch of them like if you simplify you could get 3/4.I'll give you some more.

Examples:14/16 21/24 28/32 and 35/40.

My brain was getting tired to think...but there's more.

Hope this helps :D
7 0
3 years ago
Find sum of three consecutive even integers if the largest integer is Y
KATRIN_1 [288]

Answer:

3y -- 6

Step-by-step explanation:

if the largest number is y, then the other two integers would be (y -- 2) and (y -- 4).

Therefore the sum of the integers would be

y + (y -- 2) + (y -- 4) = 3y -- 6.

7 0
3 years ago
What is the measure of the arc from A to B that does not pass through C?
elixir [45]

Answer:

a

Step-by-step explanation:

8 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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