Answer:
z = 5*(1/2)
z = 5/10
---
time switching classes:
w = 7/10
---
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (76/10 - 5/10 - 7/10)/6
x = (76 - 5 - 7)/(10*6)
x = (64)/(10*6)
x = (2*2*2*2*2*2)/(2*5*2*3)
x = (2*2*2*2)/(5*3)
x = 16/15
x = 1.0666666666
---
check:
y = 7 + 3/5
y = 7.6
z = 1/2
z = 0.5
w = 7/10
w = 0.7
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (7.6 - 0.5 - 0.7)/6
x = 1.0666666666
---
answer:
z = 5*(1/2)
z = 5/10
---
time switching classes:
w = 7/10
---
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (76/10 - 5/10 - 7/10)/6
x = (76 - 5 - 7)/(10*6)
x = (64)/(10*6)
x = (2*2*2*2*2*2)/(2*5*2*3)
x = (2*2*2*2)/(5*3)
x = 16/15
x = 1.0666666666
---
check:
y = 7 + 3/5
y = 7.6
z = 1/2
z = 0.5
w = 7/10
w = 0.7
y - 6x - z - w = 0
6x = y - z - w
x = (y - z - w)/6
x = (7.6 - 0.5 - 0.7)/6
x = 1.0666666666
---
answer:
each class is 1.07 hours
Step-by-step explanation:
Answer:34.93
Step-by-step explanation:
Using a^2+b^2=c^2we can substitute a and b in which is 34^2+8^2=c^21156+64=c^21220 = c^2Now we need to square both sides√1220 = √c^234.9284983931 ----> 34.9334.93 = cc = 34.93
Answer:
-2 is a solution
Step-by-step explanation:
-2-5<8
-7<8
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
You take 720 and divide by 4 for the 4 weeks. Then you divide by 6 for each of the day and then you reach your answer.
