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3 years ago

True or False: The hypotenuse is the longest side of a right triangle. A) yes B) no

Mathematics

1 answer:

Eva8 [605]3 years ago

4 0

Answer = A.) yes!
This statement is True
The hypotenuse is the longest side of a triangle.

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Are both students earning the same amount of money per hour? Show your work to justify your answer.

krok68 [10]

Answer:

No, Jimmy is earning $5 per hour while Amanda is earning close to $6.66 per hour.

Step-by-step explanation:

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3 years ago

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What is the range of this set of data? 111.97, 63, 84, 100, 119,72

Reika [66]

Answer:

Step-by-step explanation:

The range of a data set is the difference between the largest and the smallest values in that data set. In this case, the largest value in the data set is 119, and the smallest is 63. 119-63=56, which is the range. Hope this helps!

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3 years ago

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Find the output, d, when the input, t, is 11. d = -20+11t d= __

mafiozo [28]

Substitute the given value into the equation

-20 + 11*11

Using BODMAS, you do 11*11 first, which is 121

-20+121

Which is 101

Therefore, d is equal lot 101

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3 years ago

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Solve for the value of a in the triangle below. Round your answer to the nearest tenth

zysi [14]

Answer:

........................

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3 years ago

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

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3 years ago