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miskamm [114]
3 years ago
8

*HELP ASAP PLEASE *how can u round a decimal number to the nearest whole number?

Mathematics
2 answers:
yan [13]3 years ago
6 0

So basically the number after the decimal is not considered a whole number. A whole number is basically a number that does not have a decimal. So if this was your number, 678.710392, the whole number of that would be 679 because the number 7 after the decimal is greater than 5.

Vladimir79 [104]3 years ago
5 0

If the number (after the decimal) is 5 - 9, then add the whole number by 1.

If it is 1 - 4 only, then do not add anything to the whole number.

Example:

0.7

Since the number after the decimal is 7, and 7 is in the range of 5 - 9, then add 0 by 1.

0.7 ≈ 1.0

3.46

Since the number is 4, and is in the range of 1 - 4, then just stay it like that

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8/25, 3/6, 3/5, 0.89 is from least to greatest

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Find the six trigonometric function values of the angle θ in standard position, if the terminal side of θ is defined by x + 2y =
Black_prince [1.1K]

Answer:

\sin \theta  = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta  = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta  = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta  = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}

Step-by-step explanation:

First, we need to draw the terminal position of the given angle. To do so, we need to find a point that lies on the straight line x + 2y= 0, x\geq 0

If we choose x = 2 (we can do so because of the condition x \geq 0, which means that any positive value is suitable for x), then we have

2 +2y = 0\implies 2 = -2y \implies y = -1

Therefore, the terminal side of the angle \theta  is passing through the origin and the point  (2,-1) and now we can draw it.

The angle  \theta  is presented below.

The distance of the point  (2,-1) from the origin equals

r = \sqrt{2^2 + (-1)^2} = \sqrt{5}

Now, we can determine the values of the six trigonometric function, by using their definitions.

\sin \theta  = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta  = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta  = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta  = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}

6 0
3 years ago
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