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VLD [36.1K]
1 year ago
11

How many more barrels of gasoline then diesel were produced last year gasoline 34% diesel 29% total crude oil -60 million barrel

s Total
Mathematics
2 answers:
yarga [219]1 year ago
8 0
Diesel=20,400,000 and gasoline=17,400,000
MArishka [77]1 year ago
3 0

Answer: 3 million barrels


Step-by-step explanation:

Given :- Total crude oil= 60 million barrels

Quantity of gasoline = 34% of total crude oil

=0.34\times60\text{ million barrels}\\=20.4\text{ million barrels}

Quantity of diesel = 29% of total crude oil

=0.29\times60\text{ million barrels}\\=17.4\text{ million barrels}

Clearly, Quantity of gasoline is more than Quantity of diesel

difference = Quantity of gasoline-Quantity of diesel

20.4\text{ million barrels}-17.4\text{ million barrels}\\=3\text{ million barrels}

Therefore, The production of gasoline is 3 million barrels more than diesel last year.



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The linguists paid $9,549.65 in federal taxes.

Step-by-step explanation:

7 0
1 year ago
Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       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>h(t)=0<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:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

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 t=0.6482

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Which linear inequality is represented by the graph? y ≤ 1/2x + 2 y ≥ 1/2x + 2 y ≤ 2/3x + 2 y ≥ 2/3x + 2
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