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3 years ago

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Mathematics

2 answers:

givi [52]3 years ago

5 0

Answer:

answer A

Step-by-step explanation:

200$ to start, then 50$ a month over an unknown amount of months

Len [333]3 years ago

4 0

It’s A as that’s the only one that x is times by 50 on

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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 $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .

$\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2$ .

$\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2$ .

Solve this equation for $u$ :

$\displaystyle \frac{u^{2} + 1}{u^{2}} = 2$ .

$u^{2} + 1 = 2 u^{2}$ .

$u^{2} = 1$ .

Given that $0 \le u \le 1$ , $u = 1$ is the only possible solution.

$\cos^{2}{x} = 1$ ,

$x = k \pi$ , where $k\in \mathbb{Z}$ (i.e., $k$ is an integer.)

Given that $0 \le x < 2\pi$ ,

$0 \le k$ .

$k = 0$ or $k = 1$ . Accordingly,

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

8 0

3 years ago

Read 2 more answers

One liter is approximately equal to 0.26 gallons. Find the volume rounded to the nearest hundredth of a gallon of a container th

Leni [432]

107.2×.26= 27.87 gallons
107.2 liters, .26 gallons per liter

3 0

3 years ago

Read 2 more answers

Prove the divisibility of the following numbers:

ratelena [41]

Make use of prime factorizations:

$16^5+2^{15}=(2^4)^5+2^{15}=2^{20}+2^{15}$

Both terms have a common factor of $2^{15}$ :

$16^5+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33$

- - -

The second one is not true! We can write

$15^7+5^{13}=(3\cdot5)^7+5^{13}=3^7\cdot5^7+5^{13}$

Both terms have a common factor of $5^7$ :

$15^7+5^{13}=5^7\left(3^7+5^6\right)$

Since $30=5\cdot6$ , and $5\mid5^7$ , we'd still have to show that $5^6(3^7+5^6)$ is a multiple of 6. This is impossible, because $6=3\cdot2$ and there is no multiple of 2 that can be factored out.