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AfilCa [17]
2 years ago
7

This uses pythagorean theorem

Mathematics
2 answers:
bazaltina [42]2 years ago
8 0

Answer: the answer for rafter 1 is 13.4 and the answer for rafter 2 is 20

Step-by-step explanation: I just know

Klio2033 [76]2 years ago
7 0

Step-by-step explanation:

x²=a²+b²

x=√6²+12²

x=√180

x=3√2v

y²=16²+12²

y=√400

y=20

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For the function given state the period f(t) =6sin(3t-pi/6)-1
lapo4ka [179]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\
f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}

\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}


now, with that template in mind, let's take a peek at yours

\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}
8 0
3 years ago
How did they get that answer? please answer
OverLord2011 [107]

Answer:

Look at my explanation

Step-by-step explanation:

The circumference of a circle is 2πr, so you can set it up like this since the circumference was 22π.

2πr = 22π

divide both sides by 2

πr = 11π

divide both sides by π

you get r = 11

So your radius is 11.

To find the area, you must do πr^2 ("^2" means squared)

if you substitute 11 for r you get

11π^2

you can do 11 squared first

11^2 = 121

the area of the circle = 121π

Keep in mind that they said to use 3.14 for π

so you can do 121 x 3.14

Final Answer: 379.94

3 0
3 years ago
Plllllllllzzzzzzzzzzzzzzzzz helpppp meeeeeeeeeeeeeeeeeee
DochEvi [55]

Step-by-step explanation:

radius(CE)=diameter/2=9

EF=8inch

DE=8/2=4inch

pythagorean theorem:

im right triangle,

a^2+b^2=c^2

4^2+b^2=9^2

16+b^2=81

b^2=81-16

b^2=65

b=√65

√65≈8.06

5 0
3 years ago
An open box with a square base is to be constructed from 84 square inches of material. The height of the box is 2 inches. What a
riadik2000 [5.3K]
Your surface area is: S=x^2+4xh=84 square inches and h=2 inches; so you get:
x^2+(4×2)x=84
x^2+8x-84=0 qhich is a quadratic equation. Using the quadratic formula you get two results for x:
x1=-14 and x2=6
choose the second (positive)
6 0
3 years ago
What is the equation of the line? YA y = 2x - 4 y = {x-4 1 AX 1 y = 2x + 2 y = {x + 2​
kvasek [131]
“YA” is included in the problem?
3 0
2 years ago
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