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3 years ago

Write the following decimals as fractions. a. .76 b. .3 c. .262 d. .8053

Mathematics

2 answers:

natita [175]3 years ago

8 0

A. 19/25 b. 3/10 c. 131/500 d. 8053/10000

vampirchik [111]3 years ago

6 0

Answer: The required fractions are :

(a) $\dfrac{19}{25}.$

(b) $\dfrac{3}{10}.$

(c) $\dfrac{131}{500}.$

(d) $\dfrac{8053}{10000}.$

Step-by-step explanation: We are given to express the following decimals as fractions :

(a) : 0.76.

Here, since there are two digits after the decimal, so the fraction form is

$0.76=\dfrac{76}{100}=\dfrac{19}{25}.$

Thus, the required fraction is $\dfrac{19}{25}.$

(b) : 0.3

Here, since there is one digit after the decimal, so the required fraction form is

$0.3=\dfrac{3}{10}.$

Thus, the required fraction is $\dfrac{3}{10}.$

(c) : 0.262

Here, since there are three digits after the decimal, so the required fraction form is

$0.262=\dfrac{262}{1000}=\dfrac{131}{500}.$

Thus, the required fraction is $\dfrac{131}{500}.$

(d) : 0.8053

Here, since there are four digits after the decimal, so the required fraction form is

$0.8053=\dfrac{8053}{10000}.$

Thus, the required fraction is $\dfrac{8053}{10000}.$

The required fractions are :

(a) $\dfrac{19}{25}.$

(b) $\dfrac{3}{10}.$

(c) $\dfrac{131}{500}.$

(d) $\dfrac{8053}{10000}.$

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stira [4]

Let’s set this equation up to make this easier.

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3 years ago

General solutions of sin(x-90)+cos(x+270)=-1 {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

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