All categories

3 years ago

Match the equation with the step needed to solve it.

Mathematics

1 answer:

ivanzaharov [21]3 years ago

7 0

Send more info , you have to match the 3 with 2

You might be interested in

In triangleABC, C is the right angle. If tanA = 8/6, find cosB

klemol [59]

<u>Short answer</u> = C) 8/10

<u>Working :</u>

Tan A = 8/6 = Opposite / AdjacentOpposite = 8 units

Adjacent = 6 units

Hypotenuse = $\sqrt{8^{2} + 6^{2} }$

= $\sqrt{64 + 36 }$

= $\sqrt{100}$

= 10 units

Cos B = Adjacent/ Hypotenuse

∴ cos B= 8/10

4 0

4 years ago

When the sum of 9 and a certain number is divide by 2 the result is equal to five times the original number minus 3 what ii the

pav-90 [236]

Answer: 20/3????

Step-by-step explanation:

4 0

3 years ago

Volume of triangular prism

lukranit [14]

Answer: 225m^3

Step-by-step explanation:

Volume = Bh where "B" is the area of the base and "h" is the height

So, area of the base = 1/2(length)(height) = 1/2(5)(9) = 22.5 m^2.

Then multiply by the height.

22.5x10= 225m^3

3 0

3 years ago

Write an equation of cosine function with amplitude 2 and a period 4pi

Arisa [49]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\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}\\
\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.. hmmm let's see, keeping in mind the template above

so, amplitude of 2, A=2, thus |A| = ±2, so, you can use either

and period of 4π, well, that simply means that

$\bf \cfrac{2\pi }{B}=4\pi \implies \cfrac{2\pi }{4\pi }=B\implies \cfrac{1}{2}=B\\\\
-----------------------------\\\\

\begin{array}{llll}
f(\theta)=&\pm 2cos&\left(\frac{1}{2}\theta \right)\\
&\ \uparrow &\ \uparrow \\
&A&B
\end{array}$

8 0

4 years ago

Which equation represents the relationship shown in the table?

zalisa [80]

Answer:

How did the Silk Road connect China to the rest of the world?

Step-by-step explanation:

How did the Silk Road connect China to the rest of the world? How did the Silk Road connect China to the rest of the world? How did the Silk Road connect China to the rest of the world? How did the Silk Road connect China to the rest of the world? How did the Silk Road connect China to the rest of the world?

7 0

3 years ago