Dorrian graphed the distance, y, traveled by 2 bikes in x seconds. If the two bikes are racing a distance of 100 feet, which bik
2 answers:
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We know that
step 1
the area to install tiles is equal to A
A=area rectangle-area of a circle
area of a rectangle=(13+x)*x-----> (13x+x²) ft²
area of a circle=pi*r²-----> pi*6²------> 36*pi ft²
A=(x²+13x)-36*pi ft²
step 2
find the cost of installing tiles CT
CT=$1*[(x²+13x)-36*pi]-----> CT=$x²+$13x-$36*pi
step 3
find the cost <span>the of installing the pond CP
CP=$0.62*36*pi------> CP$22.32*pi
step 4
find the inequality
we know that
CT+CP </span>
<span> $536
[</span>$x²+$13x-$36*pi]+[$22.32*pi] $536
[$x²+$13x-$13.68*pi] $536
the answer is the option D
step 1
the area to install tiles is equal to A
A=area rectangle-area of a circle
area of a rectangle=(13+x)*x-----> (13x+x²) ft²
area of a circle=pi*r²-----> pi*6²------> 36*pi ft²
A=(x²+13x)-36*pi ft²
step 2
find the cost of installing tiles CT
CT=$1*[(x²+13x)-36*pi]-----> CT=$x²+$13x-$36*pi
step 3
find the cost <span>the of installing the pond CP
CP=$0.62*36*pi------> CP$22.32*pi
step 4
find the inequality
we know that
CT+CP </span>
[</span>$x²+$13x-$36*pi]+[$22.32*pi]
[$x²+$13x-$13.68*pi]
the answer is the option D
Answer
Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of
Step-by-step explanation:
Before we proceed let us re write correctly the height equation which in correct form reads:
Eqn(1).
Where:
: is the height range as a function of time
: is the initial velocity
: is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:
Eqn(2).
Now let us use the given information and set up our equations for Ben and Josh.
<u>Ben:</u>
We know that
Thus Eqn. (2) becomes:
Eqn.(3)
<u>Josh:</u>
We know that
Thus Eqn. (2) becomes:
Eqn. (4).
<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em><em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>
Thus we have for Ben, Eqn. (3) gives:
Using the quadratic expression to find the roots of the quadratic we have:
Since time can only be positive we reject the solution and we keep that Ben's book took
seconds to reach the ground.
Similarly solving for Josh we obtain
Thus again we reject the negative and keep the positive solution, so Josh's book took seconds to reach the ground.
Then we can find the difference between Ben and Josh times as
So to answer the original question:
<em>Whose textbook reaches the ground first and by how many seconds?</em>
- Josh's textbook reached the ground first
- Josh's textbook reached the ground first by a difference of