Can someone plzz help me with these 2 questions. thanks!

9514 1404 393

  4. height: 5√3 cm; area: 25√3 cm²

  5. 30√2 ft ≈ 42.43 ft

Step-by-step explanation:

4. The height of an equilateral triangle is (√3)/2 times the side length, so is ...

  height = (√3)/2 × (10 cm) = 5√3 cm

The area is given by the formula ...

  A = 1/2bh

  A = 1/2(10 cm)(5√3 cm) = 25√3 cm²

__

5. The diagonal of a square is √2 times the side length, so the distance from 3rd to 1st base is ...

  (30 ft)√2 ≈ 42.43 ft

__

The length of the diagonal of a square is something you should be familiar with. If you're not, you can figure the distance using the Pythagorean theorem.

  d = √(30² +30²) = √1800 ≈ 42.43 . . . feet

a is the correct

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."