Question #235875

1. Determine whether the lines given by the equations below are parallel, perpendicular, or neither. Also, find a rigorous algebraic solution for each problem.

a. {3y+4x=12 \brace -6y=8x+1}  

b.   {3y+x=12 \brace -y=8x+1}

c. {4x-7y=10 \brace 7x+4y=1}

 


2. A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by h(t)=-4.9t^2+24t+8 . What is the height of the building? What is the maximum height reached by the ball? How long does it take to reach maximum height? Also, find a rigorous algebraic solution for the problem.





3. A farmer finds that if she plants 75 trees per acre, each tree will yield 20 bushels of fruit. She estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest? Also, find a rigorous algebraic solution for the problem.



1
Expert's answer
2021-09-12T23:52:07-0400

1)

a) {3y+4x=126y=8x+1}{3y+4x=12 \brace -6y=8x+1}

For 3y+4x=123y + 4x = 12 , the slope/ gradient is 43\frac{-4}{3}

For 6y=8x+1-6y = 8x + 1 , the slope/ gradient is 43\frac{-4}{3}

Therefore, the lines are parallel.

b) {3y+x=12y=8x+1}{3y+x=12 \brace -y=8x+1}

For 3y+x=123y+x=12 , the slope/gradient is 13\frac{-1}{3}

For y=8x+1-y=8x+1 , the slope/gradient is 8-8

Therefore, the lines are neither parallel or perpendicular.

c) {4x7y=107x+4y=1}{4x-7y=10 \brace 7x+4y=1}

For 4x7y=104x-7y=10, the slope is 47\frac{4}{7}

For 7x+4y=17x+4y=1, the slope is 74\frac{-7}{4}

Therefore, the lines are perpendicular.


2)


h(t)=4.9t2+24t+8,h(t) = -4.9 t\\^2 + 24t + 8,

When t=0,h(0)=4.9(0)2+24(0)+8=8t=0, h(0) = -4.9(0)\\^2 + 24(0) + 8 = 8

Therefore, height is 8

h' = 9.8t+24-9.8t + 24

When h'=0

0=9.8t+24t=249.8=2.45sMaximum heightt=2.45,h(2.45)=4.9(2.45)2+24(2.45)+8=37.40 = -9.8 t+ 24\\t=\frac{24}{9.8} = 2.45 s\\Maximum \ height\\t=2.45,\\ h(2.45) = -4.9(2.45)\\^2 + 24(2.45) + 8 \\ = 37.4


3)

Total bushels 75×20=150075 \times 20 = 1500

Let xx be the number of tree added.

The new number is 75+x75 + x

17 bushels per tree.

Total number of bushels: =(75+x)×17=75×17+17x= (75 + x) \times 17 = 75 \times 17 + 17x

=1275+17x=1275 + 17x


=1275+17x=1275 + 17x must exceed 1500

17x17 x must exceed 225225

xx should be exceed 13.2513.25


Therefore, at least 14 trees must be added to each acre in addition to 75 in order to maximize her harvest


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