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4 years ago

Solve 8x 2 - 2x - 1 = 0 {-1/4, 1/2} {-1/2, 1/4} {1/2}

1 answer:

4 0

8x^2 - 2x - 1 = 0
(2x - 1)(4x + 1) = 0
2x - 1 = 0; x = 1/2
4x + 1 = 0;x = -1/4

answer
{-1/4, 1/2}

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3 years ago

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To model and solve our situation we are going to use the equation: $s= \frac{d}{t}$
where
$s$ is speed
$d$ is distance
$t$ is time

1. We know that the distance between the cities is 2400 miles, so $d=2400$ . We also know that the speed of the plane is 450 mi/h. Since we don't know the speed of the air, $S_{a}=?$ . We don't know how much the westward trip takes, so $t_{w}=?$ , and we also don't know how much the eastward trip takes, so $t_{e}=?$ .

Going westward. Here the plane is flying against the air, so we need to subtract the speed of the air from the speed of the plane:
$450-S_{a}= \frac{2400}{t_{w} }$
Going eastward. Here the plane is flying with the the air, so we need to add the speed of the air to the speed of the plane:
$450+S_{a}= \frac{2400}{t_{e} }$

2. We know for our problem that the round trip takes 11 hours; so the total time of the trip is 11, $t_{t}=11$ . Notice that we also know that the total time of the trip equals time of the tip going westward plus time of the trip going eastward, so $t_{t}=t_{w}+t_{e}$ . Since we know that the total trip takes 11 hours, we can replace that value in our total time equation and solve for $t_{w}$ :
$11=t_{w}+t_{e}$
$t_{w}=11-t_{e}$

Now we can replace $t_{w}$ in our going westward equation to model our round trip with a system of equations:
$450-S_{a}= \frac{2400}{t_{w}}$
$450-S_{a}= \frac{2400}{11-t_{e} }$ equation (1)
$450+S_{a}= \frac{2400}{t_{e}}$ equation (2)

3. To solve our system of equations, we are going to solve for $t_{e}$ in equations (1) (2):

From equation (1)
$450-S_{a}= \frac{2400}{11-t_{e} }$
$11-t_{e}= \frac{2400}{450-S_{a} }$
$-t_{e}= \frac{2400}{450-S_{a} } -11$
$t_{e}=11- \frac{2400}{450-S_{a} }$
$t_{e}= \frac{4950-11S_{a} -2400}{450-S_{a} }$
$t_{e}= \frac{2550-11S_{a} }{450-S_{a} }$ equation (3)

From equation (2):
$450+S_{a}= \frac{2400}{t_{e} }$
$t_{e}= \frac{2400}{450+S_{a} }$ equation (4)

Replacing (4) in (3)
$\frac{2400}{450+S_{a}} = \frac{2550-11S_{a}}{450-S_{a} }$
Now, we can solve for $S_{a}$ to find the speed of the wind:
$2400(450-S_{a})=(450+S_{a})(2550-11S_{a})$
$1080000-2400S_{a}=1147500-4950S_{a}+2550S_{a}-11(S_{a})^{2}$
$11(S_{a})^{2}-67500=0$
$11(S_{a})^{2}=67500$
$(S_{a})^{2}= \frac{67500}{11}$
$S_{a}=+/- \sqrt{ \frac{67500}{11} }$
Since speed cannot be negative, the solution of our equation is:
$S_{a}= \sqrt{ \frac{67500}{11} }$
$S_{a}=78.33$

We can conclude that the speed of the wind is 78 mph.

3 0

4 years ago