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White raven [17]
3 years ago
12

I will give brainliest to the first one to say (hey)

Mathematics
2 answers:
nirvana33 [79]3 years ago
7 0
Hey. ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Vladimir79 [104]3 years ago
5 0

Answer:

HEY! How are u? How is your day going?

Step-by-step explanation:

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The hypotenuse of a right triangle is twice the length of one of its legs. Th length of the other leg is three feet. Find the le
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Step-by-step explanation:

go for it this is your questions answer

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In a grade seven class, 18 students have pets, and 12 do not. What percent of the class has a pet?
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I need to write an even number between 7 and 16 draw a picture an them write a sentence to explain why it is an even number
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An even  number is a number that is divisible by 2.
This means that dividing this number by 2 yields no remainder (the remainder is zero).

The even numbers between 7 and 16 are:
8 as 8/2 = 4 and remainder is zero
10 as 10/2 = 5 and remainder is zero
12 as 12/2 = 6 and remainder is zero
14 as 14/2 = 7 and remainder is zero
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3 years ago
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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
GaryK [48]

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

7 0
3 years ago
Help, i've been trying for around 30mins and I can't figure it out.
omeli [17]

Answer:

<h2><em><u>Option</u></em><em><u> </u></em><em><u>A</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>C</u></em></h2>

Step-by-step explanation:

As,

2 \sqrt{75}  + 3 \sqrt{50}

= 2 \sqrt{25 \times 3}  + 3 \sqrt{25 \times 2} (A)

= 2 \times 5 \sqrt{3}  + 3 \times 5 \sqrt{2}

= 10 \sqrt{3}  + 15 \sqrt{2} (C)

6 0
3 years ago
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