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sasho [114]
2 years ago
14

Heheh be svdbtjdjsbddb

Mathematics
2 answers:
vazorg [7]2 years ago
8 0
Well he did become 3x y 6’
Dennis_Churaev [7]2 years ago
6 0
Did u type that by mistake
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Find sinθ if cosθ = <img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D" id="TexFormula1" title="\frac{2}{3}" alt="\frac{2}{
stepan [7]

Answer:

sin = \frac{\sqrt{5} }{3}

Step-by-step explanation:

                                             <u>SOH CAH TOA</u>

Where:

s = sin

c = cos

t = tan

o = opposite

a = adjacent

h = hypotenuse  

cos = 2/3 where 2 is adjacent side and 3 is hypotenuse, now find opposite side using a^2 + b^2 = c^2 //c is hypotenuse side

2^2 + b^2 = 3^2

4 + b^2 = 9

b^2 = 9 - 4

b^2 = 5

b = square root (5) = 5^(1/2) //opposite side

sin = O/H

sin = \frac{\sqrt{5} }{3}

//Hope it helps.

4 0
4 years ago
-8 ≤ 2/5(k-2)<br> What is the equation
Ksivusya [100]

i believe the answer is , k ≥ −18 , i’m not sure tho
8 0
3 years ago
Read 2 more answers
Express answer in exact form. Show all work for full credit.Find the area of one segment formed by a 6" square inscribed in a ci
Zinaida [17]

The area of the square is:

\begin{gathered} As=6^2 \\ As=36in^2 \end{gathered}

And the area of the circle is:

\begin{gathered} Ac=\pi r^2 \\ Ac=\pi\cdot(3\sqrt[]{2})^2 \\ Ac=18\pi \end{gathered}

The area of one segment is given by:

\begin{gathered} A=\frac{Ac-As}{4} \\ A=\frac{18\pi-36}{4} \end{gathered}

7 0
2 years ago
A^4 b^2÷4a^3 b^-2<br><img src="https://tex.z-dn.net/?f=%20%20%7Ba%7D%5E%7B4%7D%20%7Bb%7D%5E%7B2%7D%20%20%5Cdiv%204%20%7Ba%7D%5E%
hichkok12 [17]
A^7 /4
That is the simplification
7 0
2 years ago
Help pls i'm so behind
professor190 [17]

Answer:

I'(-3,2)

E'(-3,3)

Q'(4,4)

Z'(0,5)

6 0
3 years ago
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