Answer:
Step-by-step explanation:
solve each inequality:
A : x+6<8 , x<-8-6 , x<-14
B: x+4≥-6 , x≥-10
C: x-3>-10 , x>-7
D:x≤-9
since -12 is on the left side of the number line then x≤ -9 would be the solution
Hello,
Very nice as problem.
2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies
since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0
Answer:
Correct answer: Hmax = 22.11 m
Step-by-step explanation:
The subject of this problem is a vertical shot upwards, to which the following formula applies:
V₀² - V² = 2 · g · H where is:
V₀ - initial velocity
V - final velocity
g = 10 m/s²
H - final height
Given:
V₀ = 69 ft/s = 21.03 m/s
V = 0 m/s
Hmax = ?
V₀² = 2 · g · Hmax ⇒ Hmax = V₀² / 2 · g
Hmax = 21.03² / 2 · 10 = 442.26 / 20 = 22.11 m
Hmax = 22.11 m
God is with you!!!
Point F is on line a, so it does represent Josiah's distance at a certain time. Also, point F is below line b, so it represents a distance that is less than Chana's distance. This is a distance-time graph problem.
<h3>
What is the proof for the above?</h3>
Recall that Josiah had a head start of 10 meters and he skates at 2 meters per second.
Since Y is the function that represents the distance in meters from the finished line, by observation, it is clear to see that all the factors that are related to his race are adequately represented in:
y = 10 + 2x
Where 10 is the head start in meters
2 is the rate at which he skates per second; and
x is the unknown amount of time in seconds.
Given that the point F sits over 25 seconds,
that is F(y) = 10 + 2 * 25
= 60 meters.
Hence, Point F is on line a, so it does represent Josiah's distance at exactly 25 seconds.
Learn more about distance-time graphs at:
brainly.com/question/4931057
#SPJ1
