All categories

4 years ago

Determine the fraction for 0.54 in simplest form

Mathematics

2 answers:

ivann1987 [24]4 years ago

5 0

Answer:

27/50

Step-by-step explanation:

54 and 100 are both divisible by 2, so we get to 27/50. These have no factors in common, so 27/50 is our simplified fraction.

andrey2020 [161]4 years ago

3 0

Answer:

54/100 = 27/50

Step-by-step explanation:

You might be interested in

A jogger goes half a mile north and then turns west. If the jogger finishes 1.3 miles away from the starting point, how far west

Ratling [72]

In order to solve this problem, you must draw a right triangle first to help you visualize the path of the jogger.
You are given 0.5 miles north (which will be the height of the right triangle), and 1.3 miles (which will be the hypotenuse of the triangle).
Using Pythagorean Theorem: a²+b²=c², we have:
0.5² + b² = 1.3²
b = 1.3²-0.5²
b = 1.2 miles

3 0

3 years ago

Find f (x )if f apostrophe apostrophe (x )equals x to the power of negative 2 end exponent comma space x greater than 0 comma sp

viva [34]

Step-by-step explanation:

Given $g'(x) = g^{-2}, x> 0\ g(1) = 0 \ and \ g(2) = 0$

To get g(x), we will have to integrate g'(x)

$g(x) = \int\limits {x^{-2}} \, dx\\ g(x) = \frac{x^{-2+1}}{-2+1} + C\\ g(x) = -x^{-1} + C\\g(x) = -x^{-1} + C\\$

If g(1) = 0, this means at x = 1, g(x) = 0

0 = -1⁻¹ + C

C= 1

Substitute C = 1 into the function

g(x) = -x⁻¹ + 1

If g(2) = 0, this means at x = 2, g(x) = 0

0 = -2⁻¹ + C

C= 2⁻¹

C = 1/2

Substitute C = 2 into the function

g(x) = -x⁻¹ + 1/2

3 0

3 years ago

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$