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3 years ago

Which values are solutions to 90 < –30p + 15? check all that apply.p = –10p = 0p = –2.5p = 3p = –5p = 7.6?

Mathematics

1 answer:

Paha777 [63]3 years ago

7 0

The given inequality is

$90 < -30p + 15$

We need to isolate p, and for that , first we subtract 15 to both sides, that is

$90-15< -30p+15-15$

$75 < -30p$

Dividing both sides by -30. And on dividing by negative number, the sign of the inequality change . That is

$-2.5>p
\\
p$

And -10 is less then -2.5 . So the correct option is

$p=-10$

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Read 2 more answers

A force of 8 N makes an angle of π/4 radian with the y-axis, pointing to the right. The force acts against the movement of an ob

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

b) Θ=11.3

c) The work done is 20 $\sqrt{2}$

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)

Θ= 11.3°

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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