What are the roots of the quadratic function in the graph?
1 answer:
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Answer:
A ) Not orthogonal to each other
B) 50i + 40j + 105k
C) The tensor product is attached below
D ) The value of X = F.X is attached below
Step-by-step explanation:
attached below is the detailed solution of the above problem
A) for the vectors ( u ) and ( v ) to be orthogonal to each other [ U.V has to be = 0 ] but in this scenario U.V = 4 hence they are not orthogonal to each other
b) The vector normal to plane is gotten by : U x V
= 50i + 40j + 105k
Answer:
V=15.44
Step-by-step explanation:
We have a formula
V=\int^{π/3}_{-π/3} A(x) dx ,
where A(x) calculate as cross sectional.
We have:
Inner radius: 5 + sec(x) - 5= sec(x)
Outer radius: 7 - 5=2, we get
A(x)=π 2²- π· sec²(x)
A(x)=π(4-sec²(x))
Therefore, we calculate the volume V, and we get
V=\int^{π/3}_{-π/3} A(x) dx
V=\int^{π/3}_{-π/3} π(4-sec²(x)) dx
V=[ π(4x-tan(x)]^{π/3}_{-π/3}
V=π·(8π/3-2√3)
V=15.44
We use a site geogebra.org to plot the graph.
