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2 years ago

Team with an exame The radius of a circle is 7 kilometers. What is the length of a 45° arc?

Mathematics

1 answer:

Igoryamba2 years ago

6 0

Answer:

Step-by-step explanation:

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A particular sound wave can be graphed using the function y = 1 sin 2x y = 1 sin 2 x . Find the amplitude of the function.

alexira [117]

Answer

Explanation

The general equation of a wave is ;

y = a sin(kX+Ф)

Where a represent the amplitude, 360/k represent the period of the wave and Ф represent the phase angle.

∴ In the equation y = 1 sin 2x,

Amplitude = 1

4 0

4 years ago

Suppose a line has slope 4 and passes through the point (-2,5). Which other point must also be on the graph?

xxMikexx [17]

Slope of 4 means you add 4 to the y and 1 to the x so the answer is (-1,9)

5 0

3 years ago

In the equation y=3(4x-1)+10 , find the value of x when y = -2

kati45 [8]

Y=2(given)
-2=12x-3+10
-2=12x+7
-9=12x,
-3/4=x

6 0

4 years ago

Help I will be marking brainliest!!

Korolek [52]

Answer:

Given: base = 10.4ft, height = 12.5

Area of octagon = 8(1/2 × b × h)

8(1/2 × 10.4 × 12.5)
520ft²

Volume of pool = 520ft² × 3ft

1560ft³

Now, 1 cubic ft takes 7.5 gallons to fill.

Therefore, 1560 cubic ft takes,

1560 × 7.5
11700

So, Correct choice - [D] 11,700.

3 0

3 years ago

Help me pls! Thank you so much

frosja888 [35]

$\bf cos\left[tan^{-1}\left(\frac{12}{5} \right)+ tan^{-1}\left(\frac{-8}{15} \right) \right]\\
\left. \qquad \qquad \quad \right.\uparrow \qquad \qquad \qquad \uparrow \\
\left. \qquad \qquad \quad \right.\alpha \qquad \qquad \qquad \beta
\\\\\\
\textit{that simply means }tan(\alpha)=\cfrac{12}{5}\qquad and\qquad tan(\beta)=\cfrac{-8}{5}
\\\\\\
\textit{so, we're really looking for }cos(\alpha+\beta)$

now.. hmmm -8/15 is rather ambiguous, since the negative sign is in front of the rational, and either 8 or 15 can be negative, now, we happen to choose the 8 to get the minus, but it could have been 8/-15

ok, well hmm so, the issue boils down to

$\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad thus
\\\\\\
tan(\alpha)=\cfrac{12}{5}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}
\\\\\\
\textit{so, what is the hypotenuse "c"?}\\
\textit{ well, let's use the pythagorean theorem}
\\\\\\
c=\sqrt{a^2+b^2}\implies c=\sqrt{25+144}\implies c=\sqrt{169}\implies \boxed{c=13}\\\\
-----------------------------\\\\
\textit{this simply means }\boxed{cos(\alpha)=\cfrac{5}{13}\qquad \qquad sin(\alpha)=\cfrac{12}{13}
}$

now, let's take a peek at the second angle, angle β

$\bf tan(\beta)=\cfrac{-8}{15}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}
\\\\\\
\textit{again, let's find "c", or the hypotenuse}
\\\\\\
c=\sqrt{15^2+(-8)^2}\implies c=\sqrt{289}\implies \boxed{c=17}\\\\
-----------------------------\\\\
thus\qquad \boxed{cos(\beta)=\cfrac{15}{17}\qquad \qquad sin(\beta)=\cfrac{-8}{17}}$

now, with that in mind, let's use the angle sum identity for cosine

$\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})\\\\
-----------------------------\\\\
cos({{ \alpha}} + {{ \beta}})= \left( \cfrac{5}{13} \right)\left( \cfrac{15}{17} \right)-\left( \cfrac{12}{13} \right)\left( \cfrac{-8}{17} \right)
\\\\\\
cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}-\cfrac{-96}{221}\implies cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}+\cfrac{96}{221}
\\\\\\
\boxed{cos({{ \alpha}} + {{ \beta}})=\cfrac{171}{221}}$

8 0

3 years ago