1.Let x, y be any two numbers that satisfies the conditions x ≠0, y ≠0, and x 0 C. y/x>1

D. x/y<1

2.A pickup truck that can hold up to 3000 pounds is carrying a big machine that is 300 pounds and a few smaller ones that each weigh 60 pounds.

At least how many small machines can you fit so that it will not exceed the weight limit of the truck?

A. no more than 50

B. no less than 50

C. no less than 45

D. no more than 45

3.It usually takes Claude 40 minutes driving at 48 miles per hour to go from home to work. But due to road maintenance today, Claude has to take a detour, which makes the trip 8 miles longer than usual. What is the minimum speed Claude should travel so that he can reach the destination in less than 48 minutes?

A.30 miles per hour

B.56 miles per hour

C.50 miles per hour

D.64 miles per hour

*please make sure you answer all the questions please and thank you.

x and y can be any two numbers greater than zero such that y is also greater than x

D.no more than 45

C.50 miles per hour

Step-by-step explanation:

Let the two numbers be such that x< y because we have been given y/x>1 and x/y< 1 .

Suppose we take y= 9 and x= 3 then

9/3 > 1

3>1

Also

3/9 < 1

1/3 < 1

x and y can be any two numbers greater than zero such that y is also greater than x

2. Total weight that can be carried is 3000 pounds.

The big machine is 300 pounds. The weight that the truck can carry beside the big machine is 3000-300= 2700 pounds.

The smaller machines weigh 60 pounds

The number of smaller machines that can be carried is 2700 ÷ 60= 45 other than the big machine.

3. Total distance = Speed * time

                           = 48 * (40/60) = 32 miles

New distance = 32+ 8= 40 miles

New time = 48 minutes

Speed = distance / time = 40/ 48/60=  50 miles per hour

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

7

step-by-step explanation: