$\dfrac{1}{3}=\dfrac{1\cdot5}{3\cdot5}=\dfrac{5}{15}\\\\\dfrac{3}{5}=\dfrac{3\cdot3}{5\cdot3}=\dfrac{9}{15}\\\\\dfrac{3}{15} < \dfrac{9}{15}\ therefore\ \dfrac{3}{5}\ is\ bigger$

$Method\ 2:$

$\dfrac{1}{3} < \dfrac{1}{2}=\dfrac{1.5}{3}\\\\\dfrac{3}{5} > \dfrac{1}{2}=\dfrac{2.5}{5}\\\\therefore\ \dfrac{3}{5}\ is\ bigger$

6 0

3 years ago

Which of the following expressions are equivalent to \left(9\dfrac{7}{8} + 2\dfrac{4}{5}\right)-\dfrac{1}{2}(9 87 +2 54 )− 21

lora16 [44]

Answer:

$12\dfrac{7}{40}$

Step-by-step explanation:

The given expression is

$\left(9\dfrac{7}{8} + 2\dfrac{4}{5}\right)-\dfrac{1}{2}$

We need to find the simplified form of the given expression.

It can be rewritten as

$\left(9+\dfrac{7}{8} + 2+\dfrac{4}{5}\right)-\dfrac{1}{2}$

Combine integers and fractions separately.

$(9+2)+(\dfrac{7}{8}+\dfrac{4}{5}-\dfrac{1}{2})$

Taking LCM we get

$11+\dfrac{35+32-20}{40}$

$11+\dfrac{47}{40}$

In can be written as

$11+\dfrac{40+7}{40}$

$11+1+\dfrac{7}{40}$

$12+\dfrac{7}{40}$

$12\dfrac{7}{40}$

Therefore, the expression $12\dfrac{7}{40}$ is equivalent to the given expression.

5 0

3 years ago

Someone please help i’m desperate to pass this quiz

slavikrds [6]

Answer:

y=2x+5

Step-by-step explanation:

8 0

3 years ago

The graph shows the system ​{y=2x+1

The graph shows the system {y=2x+1