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Lostsunrise [7]
3 years ago
5

Is it possible for two equations to have more than one solution? what would that look like?​

Mathematics
1 answer:
goblinko [34]3 years ago
5 0
Its a little complicated, so I cant rlly explain it on here. But yes, 2 questions can have more than one solution. Like how theres a problem in a story, but theres plenty of solutions to solve it.
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If a escape room party has 16 participants and 4 escape puzzles.How many staff are needed?
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12

Step-by-step explanation:

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You have already traveled 25 miles. If you continue driving at the same speed, x miles per hour, for 5 more hours, you will trav
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3 years ago
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A sign is 4 1/2 feet wide and 5 1/4 feet long . Find the area of the surface of the sign.
ZanzabumX [31]

Answer: 94.5

Step-by-step explanation:

9/2 × 21/4

Get common denominator

9 × 2 =18

2 × 2 = 4

18/4 × 21/4

21 × 18 = 378

378/4 = 94.5

7 0
3 years ago
Need to know the area of the composite figure.
Deffense [45]

Answer:

  51 m^2

Step-by-step explanation:

The shaded area is the difference between the area of the overall figure and that of the rectangular cutout.

The applicable formulas are ...

  area of a triangle:

     A = (1/2)bh

  area of a rectangle:

     A = bh

  area of a trapezoid:

     A = (1/2)(b1 +b2)h

___

We note that the area of a triangle depends only on the length of its base and its height. The actual shape does not matter. Thus, we can shift the peak of the triangular portion of the shape (that portion above the top horizontal line) so that it lines up with one vertical side or the other of the figure. That makes the overall shape a trapezoid with bases 16 m and 10 m. The area of that trapezoid is then ...

  A = (1/2)(16 m + 10 m)(5 m) = 65 m^2

The area of the white internal rectangle is ...

  A = (2 m)(7 m) = 14 m^2

So, the shaded area is the difference:

  65 m^2 -14 m^2 = 51 m^2 . . . . shaded area of the composite figure

_____

<em>Alternate approach</em>

Of course, you can also figure the area by adding the area of the triangular "roof" to the area of the larger rectangle, then subtracting the area of the smaller rectangle. Using the above formulas, that approach gives ...

  (1/2)(5 m)(16 m - 10 m) + (5 m)(10 m) - (2 m)(7 m) = 15 m^2 + 50 m^2 -14 m^2

  = 51 m^2

7 0
4 years ago
Find all solutions of each equation on the interval 0 ≤ x &lt; 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
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