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2 years ago

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A particular rocket taking off from the earth surface uses fuel at a constant rate of 12.5 gallons per minute the rocket initial

ly contains 225 gallons of fuel determine a linear model and Y equals X plus B form for the amount of fuel Y as a function of the number of minutes x at the rocket has burned

1 answer:

faltersainse [42]2 years ago

3 0

Answer:

$y=12.5x+225$

Step-by-step explanation:

Let

x ----> the number of minutes

y ----> the amount of fuel in gallons

we know that

The linear equation in slope intercept form is equal to

$y=mx+b$

where

m is the unit rate or slope

b is the y-intercept or initial value

we have that

The unit rate or slope is equal to

$m=12.5\ \frac{gal}{min}$

$b=225\ gal$ ----> initial value (value of y when the value of x is equal to zero)

substitute

$y=12.5x+225$

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Answer:

Acording to the question,

Cost per cup = $1.29

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Cost of nuts = $0.69 per cup

Total number of nuts = y

Equations :

x + y = 12 (because total cups of nuts and candies will be equal to 12)

2.49x + 0.69y = 15.48 (Total cost of the 12 cups should be 15.48)

<u>Step 1 : Find x in terms of y</u>

x = 12 - y

<u>Step 2 : substitute x in terms of y from step 1 in the second equation</u>

2.49x + 0.69y = 15.48

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2 years 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

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Answer

Josh's textbook reached the ground first

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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.

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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:

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

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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
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