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4 years ago

Find two fractions with the difference of 1/5 but with neither denominator equal to 5

Mathematics

2 answers:

natulia [17]4 years ago

6 0

The two fractions with the difference of 1/5 but with neither denominator equal to 5 is 4/10 and 2/10.

lidiya [134]4 years ago

3 0

4/10 and 2/10 is the answer I believe

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X- (2x + 1) = 8- (3x + 3)

maxonik [38]

Answer:

x = 3

Step-by-step explanation:

x- (2x + 1) = 8- (3x + 3)

x - 2x - 1 = 8 - 3x - 3

x - 2x + 3x = 8 - 3 + 1

2x = 6

x = 6/2 = 3

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3 years ago

Is 0.333 a natural number?

NeX [460]

Answer:

No, 0.333 is a rational number, not a natural number

Step-by-step explanation:

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3 years ago

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Write an equation in standard form for the line that passes through the given points.

Agata [3.3K]

So, pretend this is your x-axis and y-axis:

(-2,7) • I

I • (2, 5)

_________________I____________________

TO GET FROM POINT (-2, 7) TO POINT (2, 5), WE MOVE DOWN 2 AND OVER 4, SO THE SLOPE IS -1/2. IF WE FOLLOW THAT SLOPE AND MOVE DOWN 1 AND OVER 2 FROM THE FIRST POINT OF (-2, 7), WE WILL LAND ON A POINT LOCATED AT (0, 6), WHICH WOULD BE THE "Y-INTERCEPT". WE WERE JUST ABLE TO CALCULATE THE SLOPE OF THE LINE AND THEN USE THE SLOPE TO FIND THE INTERCEPT. SO, THE "SLOPE-INTERCEPT" FORM OF THE EQUATION FOR THIS LINE IS:

y = -1/2x + 6

TO RE-WRITE THIS IN STANDARD FORM, WE JUST WANT TO MOVE THE X VARIABLE OVER TO THE LEFT WITH THE Y VARIABLE, SO:

y = -1/2x + 6

+1/2x + 1/2x

1/2x + y = 6 .... and that is your answer!

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2 years ago

What is 50 divided by 2/3

katovenus [111]

Answer:

Step-by-step explanation:

6 0

4 years ago

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Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$

5 0

1 year ago