Solve the equation below for x. -1 2(3x - 4) = 11

2 answers:

7 0

If the -1 is not included, then your answer is 3.17.

2(3x-4)=11

3x2=6 and -4x2= -8

6x-8=11

+8 +8

6x=19

/6 /6

X= 3.17

vovangra [49]3 years ago

3 0

If the -1 not included, then your answer is 3.17.

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The arch beneath a bridge is semi-elliptical, a one-way roadway passes under the arch. The width of the roadway is 38 feet and

forsale [732]

Answer:

Only truck 1 can pass under the bridge.

Step-by-step explanation:

So, first of all, we must do a drawing of what the situation looks like (see attached picture).

Next, we can take the general equation of an ellipse that is centered at the origin, which is the following:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}$

where:

a= wider side of the ellipse

b= shorter side of the ellipse

in this case:

$a=\frac{38}{2}=19ft$

and

b=12ft

so we can go ahead and plug this data into the ellipse formula:

$\frac{x^2}{(19)^2}+\frac{y^2}{(12)^2}$

and we can simplify the equation, so we get:

$\frac{x^2}{361}+\frac{y^2}{144}$

So, we need to know if either truk will pass under the bridge, so we will match the center of the bridge with the center of each truck and see if the height of the bridge is enough for either to pass.

in order to do this let's solve the equation for y:

$\frac{y^{2}}{144}=1-\frac{x^{2}}{361}$

$y^{2}=144(1-\frac{x^{2}}{361})$

we can add everything inside parenthesis so we get:

$y^{2}=144(\frac{361-x^{2}}{361})$

and take the square root on both sides, so we get:

$y=\sqrt{144(\frac{361-x^{2}}{361})}$

and we can simplify this so we get:

$y=\frac{12}{19}\sqrt{361-x^{2}}$

and now we can evaluate this equation for x=4 (half the width of the trucks) so:

$y=\frac{12}{19}\sqrt{361-(8)^{2}}$

y=11.73ft

this means that for the trucks to pass under the bridge they must have a maximum height of 11.73ft, therefore only truck 1 is able to pass under the bridge since truck 2 is too high.