Answer:
24
Step-by-step explanation:
1250/ 54 = 23.14814814
Answer:
a) F= 4
i + 4
j
b) Θ=11.3
c) The work done is 20
Step-by-step explanation:
a) ||F||=8, α=π/4
Fx=||F||·sin(π/4)=8·
Fy=||F||·cos(π/4)=8·
F=Fx i + Fy j = 4
i + 4
j
b) We can find the value of Θ using the equation:
cos(Θ)=
where:
D= 3 i + 2 j
F=4
i + 4
j
The dot product is defined as the sum of the products of the components of each vector as:
F · D= 
||F||= 8
||D||= 
Hence:
Θ=arccos(
)
Θ=arccos(0.981)
Θ= 11.3°
c) Work is equal to:
F · D=
=28.3
Other way of obtainig the work is:
||F||||D||cos(Θ)
where:
||F||= 8
||D||= 
Θ=11.3°
So, ||F||||D||cos(Θ)=8×
×cos(11.3°)=28.3
Answer:
19 m
Step-by-step explanation:
Perimeter = 2(length + width) ; P = 2(l+w) - - (1)
Area = Length * width ; A = l*w - - - (2)
76 = 2(l+w)
76/2 = l+w
l+w = 38
l = 38 - w
Put l = 38 - w in (1)
A = (38-w)*w
A = 38w - w²
At maximum point:
dA/dw = 0
dA/dw = 38 - 2w
38 - 2w = 0
38 = 2w
w = 38/2
w = 19
What are you asking me ? Do you have a picture I can look at ?
If your question looks like mine (shown in picture).Your answer would be number 4.
Hope this helps!
CTPehrson