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blagie [28]
2 years ago
11

HELP QUICK i need to turn this in in 20 mins

Mathematics
1 answer:
tia_tia [17]2 years ago
4 0
25 for the PS ARC ANGLE
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Can Someone Answer My Final Answers :) I’d appreciate it so much :)
DaniilM [7]
1. Remember that the perimeter is the sum of the lengths of the sides of a figure.To solve this, we are going to use the distance formula: d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}
where
(x_{1},y_{1}) are the coordinates of the first point
(x_{2},y_{2}) are the coordinates of the second point
Length of  WZ:
We know form our graph that the coordinates of our first point, W, are (1,0) and the coordinates of the second point, Z, are (4,2). Using the distance formula:
d_{WZ}= \sqrt{(4-1)^2+(2-0)^2}
d_{WZ}= \sqrt{(3)^2+(2)^2}
d_{WZ}= \sqrt{9+4}
d_{WZ}= \sqrt{13}

We know that all the sides of a rhombus have the same length, so 
d_{YZ}=  \sqrt{13}
d_{XY}= \sqrt{13}
d_{XW}= \sqrt{13}

Now, we just need to add the four sides to get the perimeter of our rhombus:
perimeter= \sqrt{13} + \sqrt{13} + \sqrt{13} + \sqrt{13}
perimeter=4 \sqrt{13}
We can conclude that the perimeter of our rhombus is 4 \sqrt{13} square units. 

2. To solve this, we are going to use the arc length formula: s=r \alpha
where
s is the length of the arc. 
r is the radius of the circle.
\alpha is the central angle in radians

We know form our problem that the length of arc PQ is \frac{8}{3}  \pi inches, so s=\frac{8}{3} \pi, and we can infer from our picture that r=15. Lest replace the values in our formula to find the central angle POQ:
s=r \alpha
\frac{8}{3} \pi=15 \alpha
\alpha =  \frac{\frac{8}{3} \pi}{15}
\alpha = \frac{8}{45} \pi

Since \alpha =POQ, We can conclude that the measure of the central angle POQ is \frac{8}{45} \pi

3. A cross section is the shape you get when you make a cut thought a 3 dimensional figure. A rectangular cross section is a cross section in the shape of a rectangle. To get a rectangular cross section of a particular 3 dimensional figure, you need to cut  in an specific way. For example, a rectangular pyramid cut by a plane parallel to its base, will always give us a rectangular cross section. 
We can conclude that the draw of our cross section is:

6 0
3 years ago
Read 2 more answers
Find a particular solution to y" - y + y = 2 sin(3x)
leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

8 0
2 years ago
The sum of three consecutive terms of an arithmetic sequence is 27, and the sum of their squares is 293. What is the absolute di
Ber [7]

The absolute difference between the greatest and the least of these three numbers in the arithmetic sequence is 10.

The sequence is an arithmetic sequence. Therefore,

d = common difference

let

a = centre term

Therefore, the 3 consecutive term will be as follows

a - d,  a, a + d

a - d +  a + a + d = 27

3a = 27

a = 27 / 3

a = 9

Therefore,

(a-d)² + (a)² + (a + d)² = 293

(a²-2ad+d²) + 9² + (a² + 2ad + d²) = 293

(81 - 18d + d²) + 81 + (81 + 18d + d²) = 293

243 + 2d² = 293

2d² = 50

d² = 50 / 2

d = √25

d = 5

common difference = 5

Therefore, the 3 numbers are as follows

9 - 5 , 9, 9 + 5 = 4, 9, 14

The difference between the greatest and the least of these 3 numbers are as follows:

14 - 4 = 10

learn more on Arithmetic progression: brainly.com/question/25749583?referrer=searchResults

7 0
2 years ago
8(y+4)-2(y-1)=70-3y help pls
jeka57 [31]

Answer:

y = 4

Step-by-step explanation:

8(y+4)-2(y-1)=70-3y

8y+32-2y+2=70-3y

Grouping of like terms

8y-2y+3y=70-32-2

9y=36

Divide both sides by 9

9y/9 = 36/9

9 cancels itself and 9.

y=36/9

y=4

Therefore, y = 4.

Hope it helps.

If so, mark as brainliest.

8 0
3 years ago
Read 2 more answers
3. Solve the system using elimination (not substitution or matrices). negative 2 x plus y minus 2 z equals negative 8A N D7 x pl
riadik2000 [5.3K]

Elimination Method

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ 5X+2Y-Z=-9 \end{gathered}

If we multiply the equation 3 by (-1) we obtain this:

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ -5X-2Y+Z=9 \end{gathered}

If we add them we obtain 0, therefore there are infinite solutions. So, let's write it in terms of Z

1. Using the 3rd equation we can obtain X(Y,Z)

\begin{gathered} 5X=-9-2Y+Z \\ X=\frac{-9-2Y+Z}{5} \\  \end{gathered}

2. We can replace this value of X in the 1st and 2nd equations

\begin{gathered} -2\cdot(\frac{-9-2Y+Z}{5})+Y-2Z=-8 \\ 7\cdot(\frac{-9-2Y+Z}{5})+Y+Z=-1 \end{gathered}

3. If we simplify:

\begin{gathered} \frac{-9Y+12Z-63}{5}=-1 \\ \frac{9Y-12Z+18}{5}=-8 \end{gathered}

4. We can obtain Y from this two equations:

\begin{gathered} Y=-\frac{-12Z+58}{9} \\  \end{gathered}

5. Now, we need to obtain X(Z). We can replace Y in X(Y,Z)

\begin{gathered} X=\frac{-9-2Y+Z}{5} \\ X=\frac{-9-2(-\frac{-12Z+58}{9})+Z}{5} \end{gathered}

6. If we simplify, we obtain:

X=\frac{-3Z+7}{9}

7. In conclusion, we obtain that

(X,Y,Z) =

(\frac{-3Z+7}{9},-\frac{-12Z+58}{9},Z)

8 0
11 months ago
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