3 years ago

Let f(x) = x^2 +3x and g(x) =2x -1 (f+g)(3)

Mathematics

1 answer:

Rainbow [258]3 years ago

3 0

let f=1

then

1^2+3(1)

2(1)-1(5+1)(3)=19

hope this is the answer plz mark brillent thank you my friend

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Answer:

a) F= 4 $\sqrt{2}$ i + 4 $\sqrt{2}$ j

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Step-by-step explanation:

a) ||F||=8, α=π/4

Fx=||F||·sin(π/4)=8· $\frac{\sqrt{2}}{2}$

Fy=||F||·cos(π/4)=8· $\frac{\sqrt{2}}{2}$

F=Fx i + Fy j = 4 $\sqrt{2}$ i + 4 $\sqrt{2}$ j

b) We can find the value of Θ using the equation:

cos(Θ)= $\frac{F.D}{||F||||D||}$

where:

D= 3 i + 2 j

F=4 $\sqrt{2}$ i + 4 $\sqrt{2}$ j

The dot product is defined as the sum of the products of the components of each vector as:

F · D= $(4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}$

||F||= 8

||D||= $\sqrt{3^2+2^2} =\sqrt{13}$

Hence:

Θ=arccos( $\frac{20\sqrt{2} }{8\sqrt{13} }$ )

Θ=arccos(0.981)

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c) Work is equal to:

F · D= $(4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}$ =28.3

Other way of obtainig the work is:

||F||||D||cos(Θ)

where:

||F||= 8

||D||= $\sqrt{3^2+2^2} =\sqrt{13}$

Θ=11.3°

So, ||F||||D||cos(Θ)=8× $\sqrt{13}$ ×cos(11.3°)=28.3

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