#3 only Fractions help

Mathematics

2 answers:

aalyn [17]3 years ago

6 0

Answer:

Step-by-step explanation:

24/1/3 -8/5/6- 8/1/2 (accorin to fractions la)

improper fraction make it into proper fraction

73/3- 53/6- 17/2 (change the base of 3 and 2 into 6)

146/6-53/6-51/6= (146-53-51)/6

= 7

pantera1 [17]3 years ago

5 0

Answer:

Step-by-step explanation:

you got to try your hardest

You might be interested in

The Russo-Japanese War was a conflict between Russia and Japan that started in the year 1904. Let x represent any year. Write an

sladkih [1.3K]

The years when the Russo-Japanese war had not yet happened is the year of 1903 and before

Let x represents years, the inequality system is x < 1904

6 0

3 years ago

pirates 1.05 * 2.8 by calculating 105 * 28 then moving to decimal 3 places to the left Why does Priya method make sense

Mnenie [13.5K]

Answer:

YES CORRECT BOI

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$