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:

You might be interested in

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

shutvik [7]

Step-by-step explanation:

go for it this is your questions answer

3 0

3 years ago

In a grade seven class, 18 students have pets, and 12 do not. What percent of the class has a pet?

Svetradugi [14.3K]

Answer :
If 18+12 is 30 and only 18 of them has pets then 70 percent have pets

7 0

3 years ago

Read 2 more answers

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

serious [3.7K]

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

5 0

3 years ago

Read 2 more answers

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

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