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telo118 [61]
3 years ago
10

The sum of the squares of two positive numbers is 41 and the difference of the two squares of the numbers is 9. Find the numbers

. If there is more than one pair, use the "or" button.
Mathematics
1 answer:
vekshin13 years ago
8 0

Answer:

Step-by-step explanation:

41

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Is this the correct answer?
charle [14.2K]
Yes, it is right... you correctly distributed the negative, and combined like terms... great job!
6 0
3 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.

\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
2 years ago
Read 2 more answers
Solve the inequality.<br><br> y−5≥8
mixas84 [53]

Answer:

Y is less than or equal to 13

Step-by-step explanation:

3 0
3 years ago
What is 3 divided by 4/5
Veseljchak [2.6K]
The answer is 3.75 or 3 3/4
3 0
2 years ago
Read 2 more answers
Dos cuadrados de lado
Kaylis [27]

La franja amarilla del rectángulo tiene un área de 30 centímetros cuadrados.

<h3>¿Cuál es el área de la franja amarilla del rectángulo?</h3>

En este problema tenemos un rectángulo formado por dos cuadrados que se traslapan uno al otro. La franja amarilla es el área en la que los cuadrados se traslapan. La anchura del rectángulo es descrita por la siguiente ecuación:

(10 - x) + 2 · x = 17

Donde x se mide en centímetros.

A continuación, despejamos x en la ecuación descrita:

10 + x = 17

x = 7

Ahora, el área de la franja amarilla se determina mediante la fórmula de area de un rectángulo:

A = b · h

Donde:

  • b - Base del rectángulo, en centímetros.
  • h - Altura del rectángulo, en centímetros.
  • A - Área del rectángulo, en centímetros cuadrados.

A = (10 - 7) · 10

A = 3 · 10

A = 30

El área de la franja amarilla del rectángulo es igual a 30 centímetros cuadrados.

Para aprender más sobre áreas de rectángulos: brainly.com/question/23058403

#SPJ1

8 0
11 months ago
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