Please help, thank you! I'll give brainliest to the correct answer with an explanation.

O is the center of both circles containing arcs AB and CD

victus00 [196]

Answer:

Step-by-step explanation:

You need to solve the area sector of the bigger circle. Since we now the length of sector is 270, we can set up an equation.

Length of sector DC = 2 $\pi$ r

270 = 2 $\pi$ (8+OA) *(8+OA is the raidus of the bigger circle/sector)*

270/2 $\pi$ = 8 + OA (division)

135/ $\pi$ -8 = OA

135/ $\pi$ -8 = OB (because OA and OB are the raidus of the same circle, so they are congruent)

I hope this helps you! Please correct me if you see a miscalculation <3<3

5 0

3 years ago

In a parking lot there are totally 70 vehicles, cars and motorcycles. all together they've got 258 wheels. how many cars are the

docker41 [41]

1 car - 4 wheels
1 motorcycle - 2 wheels

x - number of cars
y - number of motorcycles

so, there are 70 vehicles: x + y = 70
and all togeather they've got 258 wheels: 4*x + 2*y = 258
because each car has 4 wheels, so all car wheels = 4*x,
likewise 2*y for motorcycles

x + y = 70 /* 4
4 x + 2y = 258

4 x + 4 y = 280 (R1)
4 x + 2y = 258 (R2)

2y = 22 /:2 it is R1 - R2
x = 70 - y

y = 11 /motorcycles
x = 70 - 11 = 59 /cars

check:
59 + 11 = 70 ok
59 * 4 + 11 * 2 = 236 + 22 = 258 ok

5 0

3 years ago

Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g

SpyIntel [72]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see $\bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad B
\end{array}$

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.

8 0

3 years ago

Read 2 more answers

What is -16-15????????????

frozen [14]

Answer:

1

Step-by-step explanation:

to lazy

6 0

3 years ago

Read 2 more answers