CL.1.142):
A):
x5 ==> 4 boxes = 3ft <==== x5
20 boxes = ? ft
20 boxes is 15 ft. high
B): x3 ===> 4 boxes = 3ft. <==== x3
? boxes = 9ft.
12 boxes will fit in one stack.
c): CL.143
Perimeter = 15 + 29 + 9 + 11 + 6 + 18 =====> 88m
A1 = b * h
(18)(15)
270m^2
A2 = b * h
(11)(9)
99
Total Area ======> 270 + 99 ======> 369m^2
CL.1-144
1/4 + 1/4 + 1/5 = 5/20 + 5/20 + 4/20 ====> 14/20
1/4 * 5/5 =====> 5/20
1/5 * 4/4 =====> 4/20
14/20 + ?/20 = 20/20
? = 6 Missing Section
6/20 ======> 3/10
CL-1.145
A): 40/100 = 4/10 ====> 2/5
0.4 ========> 40%
B): 1/6 ======> .00000
0.16 (Repeating 6) ======> 16.6%
C): 37.5/100 = 375/1000 =======> 3/8
0.375 =========> 37.5%
CL-1.146
ADD: 1/6 + 1/2 show all steps:
1/6 + 1/2
1/6 + 3/6
1 + 3 / 6
4/6 =======> 2/3 =======> Decimal: =======> 0.666667
Hope that helps!!!! : )
A linear function is y=mx + b so yes (assuming + sign at end is accidental)
Answer:
The stone will be in flight for 3 seconds
Step-by-step explanation:
Given the equation;
d = -16t^2 + vt + h
Where;
v = the initial velocity
h = the initial height
Given;
v = 48 ft/s
h = 0
Substituting into function d;
d = -16t^2 + 48t + 0
d = -16t^2 + 48t
At the point of launch and the point of landing d = 0.
We need to calculate the values of t for d to be equal to zero.
d = -16t^2 + 48t = 0
-16t^2 + 48t = 0
Factorising, we have;
16(-t+3)t = 0
So,
t = 0 or -t+3 = 0
t= 0 or t =3
Change in t is;
∆t = t2 - t1 = 3 - 0
∆t = 3 seconds.
The stone will be in flight for 3 seconds.
Answer:
50.24cm
Step-by-step explanation:
C = 2πr
C = 2(3.14)8
C = 3.14(16)
C = 50.24 cm
Never heard of “zinc” sorry buddy cant help
