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3 years ago

Which expression can be used to convert 22 Australian dollars to US dollars?

Mathematics

2 answers:

krek1111 [17]3 years ago

7 0

Answer:

18.3 USD

Step-by-step explanation:

Because 1.2 AUD = 1 USD

Therefore, $22 AUD / 1.2 = USD

22 * 1.2 ≈ 18.3 USD

Hope this help :3

kotegsom [21]3 years ago

5 0

Answer:

(22/1.2)AUD = 18.33 USD

Step-by-step explanation:

Given

1.2 Australian dollars = 1 US Dollar

To convert 22 Australian dollars to US dollars we will divide 22 by 1.2.

Therefore,

22 Australian dollars = $\frac{22}{1.2}$

= 18.33 USD

The expression can be used

(22/1.2)AUD = 18.33 USD

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Please Help? A circle has an arc length of 5π in. The central angle for this arc measures π/3 radians. What is the area of the a

Zielflug [23.3K]

Answer:

The area of the associated sector is $\frac{25}{24}\pi \ in^{2}$

Step-by-step explanation:

step 1

Find the radius of the circle

we know that

The circumference of a circle is equal to

$C=2\pi r$

we have

$C=5\pi\ in$

substitute and solve for r

$5\pi=2\pi r$

$r=2.5\ in$

step 2

Find the area of the circle

we know that

The area of the circle is equal to

$A=\pi r^{2}$

we have

$r=2.5\ in$

substitute

$A=\pi (2.5^{2})=6.25\pi\ in^{2}$

step 3

Find the area of the associated sector

we know that

$2\pi\ radians$ subtends the complete circle of area $6.25\pi\ in^{2}$

by proportion

Find the area of a sector with a central angle of $\pi/3\ radians$

$\frac{6.25\pi }{2\pi} =\frac{x}{\pi/3}\\x=6.25*(\pi/3)/2\\ \\x=\frac{25}{24}\pi \ in^{2}$

8 0

3 years ago

QUICK !!!

Alexxx [7]

Looks to me like y= x + 1

5 0

2 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .