Answer: 11.25 secs
Step-by-step explanation:
So in this sense the rockets origin or the ground is modeled at h=0 so the time required if used on a table shows that h=0 between the values of 11 and 12. So if you plug and chug decimal values between these two values you get exactly 0 at t=11.25 so it takes approximately 11.25 seconds for the rocket to return to the ground
You just want to simplify right?!
45. (a^2b^3)(ab)^-2
= (a^2b^3)(a^-2b^-2)
= b
46. (-3x^3y)^2(4xy^2)
= (-9x^6y^2)(4xy^2)
= -36x^7y^4
47. 3c^2d(2c^3d^5) / 15c^4d^2
= 6c^5d^6 / 15c^4d^2
= 2/5c1/4x^4
48. -10g^6h^9(g^2h^3) / 30g^3h^3
= -10g^8h^12 / 30g^3h^3
= -1/3g^5h^9
49. 5x^4y^2(2x^5y^6) / 20x^3y^5
= 10x^9y^8 / 20x^3y^5
= 1/2x^6 1/3y^3
50. -12n^7p^5(n^2p^4) / 36n^6p^7
= -12n^9p^9 / 36n^6p^7
= -1/3n^3p^2
(Sorry it’s messy it’d look better if my phone could actually put the numbers to the power)
If in 2 hours he read 200 pgs then in 1 hour will be 50 pgs
Answer:
B. Hope this helps
Step-by-step explanation:
Answer:
y = 65x
Step-by-step explanation:
