I need help with these three problems please

Q7. 11.3 inches (3 s.f.)

Q8. 96.2 ft

Q9. 36.4cm

Step-by-step explanation:

Q7. Please see attached picture for full solution.

Q8. Let the length of a side of the square be x ft.

Applying Pythagoras' Theorem,

Thus, the perimeter of the square is

Q9. Equilateral triangles have 3 equal sides and each interior angle is 60°.

Since the perimeter of the equilateral triangle is 126cm,

length of each side= 126÷3 = 42 cm

The green line drawn in picture 3 is the altitude of the triangle.

Let the altitude of the triangle be x cm.

sin 60°=

(to 3 s.f.)

Therefore, the length of the altitude of the triangle is 36.4cm.

7) 8sqrt(2) in

8) 68sqrt(2) ft

9) 21sqrt(3) cm

Step-by-step explanation:

7) sqrt(8² + 8²) = 8sqrt(2) in

sqrt is square root

8) sqrt(x² + x²) = 34

2x² = 34²

x = 17sqrt(2)

Perimeter = 4x = 68sqrt(2) ft

9) 3s = 126

s = 42

Altitude² + 21² = 42²

Altitude² = 1323

Altitude = 21sqrt(3) cm

The question is incomplete. Here is the complete question:

Alek went for a walk. D(t) models the distance Alek walked (in kilometers) after t hours. What does the statement D(0.5) < D(1) - D(0.5) mean?

It means that Alex traveled more distance during the second half hour than the first.

Step-by-step explanation:

Given:

The distance traveled by Alex after 't' hours = where 'D' is in km.

D(0.5) < D(1) - D(0.5).

Now, D(0.5) means distance traveled in 0.5 hours = 'x' (Let)

D(1) means distance traveled in 1 hour = 'y' (Let)

Now, D(1) - D(0.5) =

Thus, D(1) - D(0.5) is the difference between the distance traveled after 1 hour and after 0.5 hours.

So, this is nothing but the distance traveled in the second half hour.

Now, it is given that, D(0.5) < D(1) - D(0.5).

So, distance traveled in the first half hour is less than the distance traveled in the next half hour or Alex traveled more distance during the second half hour than the first.