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fiasKO [112]
3 years ago
7

What is-2x+6y=6 7x+8y=-6?

Mathematics
1 answer:
rodikova [14]3 years ago
3 0
We can solve this by substitution

-2x=6-6y

-2x/-2=(6-6y)/-2
x= -3+3y

7(-3+3y)+8y=-6
-21+21y+8y=-6
29y=15
y=15/29

-2x+6(15/29)=6
-2x+90/29=6
-2x=6-90/29
-2x= 174/29-90/29
-2x= 84/29
-2x(29)=(84/29)29
-54x=84
x= -84/54
x= -14/9

so x= -14/9
y= 15/29
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The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

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a_2 the horizontal distance between the second observer and the ballonist

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S=a*h (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

S=S_1+S_2 (equation 2)

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But we can write S_1 in terms of \alpha _1, like this:

\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}

And for S_2 will be the same:

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Replacing in the equation 2:

S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})

And replacing in the equation 1:

h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}

So, we can replace all the known data in the last equation:

h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

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