Let $v_0$ be the cat's speed just as it leaves the edge of the table. Then taking the point 1.3 m below the edge of the table to be the origin, the cat's horizontal position at time $t$ is given by

$x(t)=v_0t$

and its height is

$y(t)=1.3\,\mathrm m-gt^2$

where $g$ is 9.8 m/s^2, the magnitude of the acceleration due to gravity.

The time it takes for the cat to hit the ground is $t$ with

$0=1.3\,\mathrm m-gt^2\implies t=\sqrt{\dfrac{1.3\,\rm m}g}\approx0.36\,\mathrm s$

(Unfortunately, this doesn't match any of the given options...)

The cat lands 0.75 m away (horizontally) from the edge of the table, so that its speed $v_0$ was

$0.75\,\mathrm m=v_0(0.36\,\mathrm s)\implies v_0\approx2.08\dfrac{\rm m}{\rm s}$

(Again, not one of the answer choices...)

I'm guessing there's either a typo in the question or answers.

6 0

2 years ago

Which statements are TRUE? Select all that apply.

musickatia [10]

Answer:

A, B, D are true

Step-by-step explanation:

A) Speed = Distance travelled ÷ time taken

= 112 ÷ 1.75

= 64 miles per hour

True

B) Cost of 6 pens = $7.74

Cost of 1 pen = 7.74 ÷ 6 = $ 1.29

Cost of 7 pens = 1.29 * 7 = $ 9.03

TRUE

C) Speed = 57 miles per hour

4 hour 10 minutes = 250 minutes

Distance traveled in an hour = 57

Distance traveled in 250 minutes = (57* 250) ÷ 60 = 237.50 miles

FALSE

D) Cost of 17 cans = $ 38.59

Cost of 1 can = 38.59 ÷ 17 = $2.27

Cost of 3 cans = 2.27 * 3 = $ 6.81

TRUE

e) Cost of two dozen eggs = $ 8.40

Cost of an egg = 8.40 ÷ 24 = $ 0.35

FALSE

7 0

2 years ago