All categories

3 years ago

Round 471.42857 to the nearest cent

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

6 0

Answer:

471

Step-by-step explanation:

do to the number in the tenths place not being higher than 5 all number past the decimal become zero and the whole number stay the same

You might be interested in

(Picture) INFINITE SEQUENCES AND SERIES PLEASE HELP!!

mojhsa [17]

Answer:

C) $x=\frac{17}{33}$

Step-by-step explanation:

0.51 repeating decimal

x= 0.51515151

multiply by 100 on both sides

100x = 51.515151.........

x= 0.515151...............

--------------------------------------------(subtract)

99x = 51

divide by 99 on both sides

$x=\frac{51}{99}$

divide both sides by 3

$x=\frac{17}{33}$

5 0

3 years ago

The perimeter of a square is 1.2m. Find the length of the side.<br> The area of the square.

Nana76 [90]

<span>length of the side = 1.2 / 4 = 0.3 m

area = 0.3</span>² = 0.09 m²

8 0

4 years ago

Read 2 more answers

Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p

kirza4 [7]

The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

$\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)$

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>

In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
In second quadrant(π/2 < θ < π), only sin and cosec are positive.
In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

$y= 3\cos(2(x + \pi/2)) - 2$

The options are:

$y= 3\sin(2(x + \pi/4)) - 2$
$y= -3\sin(2(x + \pi/4)) - 2$
$y= 3\cos(2(x + \pi/4)) - 2$
$y= -3\cos(2(x + \pi/2)) - 2$

Checking all the options one by one:

Option 1: $y= 3\sin(2(x + \pi/4)) - 2$

$y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2$

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

Option 2: $y= -3\sin(2(x + \pi/4)) - 2$

This option if would be true, then from option 1 and this option, we'd get:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$