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3 years ago

Type your number answer in decimal form. do not round . ? millimeters = 9.2 cenimeters

Mathematics

2 answers:

alekssr [168]3 years ago

8 0

Answer:

0.92 centimeter

Step-by-step explanation:

1 centimeter contain 10 millimeter

1 millimeter=1/10=0.1 centimeter

9.2 millimeter= 9.2/10=0.92

Good luck Please mark me brainliest

adell [148]3 years ago

3 0

Answer:

10 centimeter=92 milimeter

Step-by-step explanation:

we know that

one centimeter contains 10 milimeter or

1 centimeter is equal to 10 milimeter

1 centimeter=10 milimeter

we have to find

how many milimetersare there in 9.2 milimeters

9.2 centimeter=? milimeter

for this we will multiply by 9.2 on both sides

1 centimeter=10 milimeter

9.2 centimeter=10*(9.2)=92 milimeter

10 centimeter=92 milimeter

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Step-by-step explanation:

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Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

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Answer:

d=10u

Q(5/3,5/3,-19/3)

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:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

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

Substitute the value of $\lambda$ in the parametric equations:

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

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

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

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