$\dfrac{1}{7}-3\left(\dfrac{3}{7}n-\dfrac{2}{7}\right)=\dfrac{1}{7}-\dfrac{3\cdot3}{7}n+\dfrac{3\cdot2}{7}=\dfrac{1}{7}+\dfrac{6}{7}-\dfrac{9}{7}n\\\\=\dfrac{1+6}{7}-\dfrac{9}{7}n=\boxed{1-\dfrac{9}{7}n}$

5 0

2 years ago

Read 2 more answers

Math 170

Alinara [238K]

The given sides are 12 cm, 15 cm and 21 cm

Shortest side is 12 cm

As given that in real the shortest side is 200 m

So 12 cm=200 m

then 1 cm = $\frac{200}{12}=\frac{50}{3}$

Hence 15 cm = $\frac{50}{3} * 15 = 250$ m

21 cm = $\frac{50}{3} * 21 = 350$ m

So,,

15 cm in the photograph is 250 m long .

21 cm in the photograph is 350 long.

8 0

3 years ago

According to the text, which element has the LEAST number of dimensions? A figure B tine C) plane D) point

Anvisha [2.4K]

Point is the least number of dimensions

Im sure :)

8 0

3 years ago