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2 years ago

Michael is 3 times as old as Brandon. 18 years ago, Michael was 9 times as old as Brandon.

Mathematics

1 answer:

Mrrafil [7]2 years ago

5 0

Answer: Brandon is 24, Michael is 72.

Step-by-step explanation:

Let m represent Michael

Let b represent Brandon

Then,

m = 3b <-----------"Michael is 3 times as old as Brandon."

m - 18 = 9(b - 18) <-----------"18 years ago, Michael was 9 times as old as Brandon."

3b - 18 = 9b - 162

-18 = 6b - 162

144 = 6b

Michael is 3 times as old as Brandon. Also, 18 years ago, Brandon would have been 6 and Michael would have been 54 meaning Michael was 9 times as old as Brandon 18 years ago.

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Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

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$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

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$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

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$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

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