Question

MEDAL!!!


Represent the area of a triangle with a base of
2x + y and a height of 4z.

A. 2z(2x + y)
B. 2x( z + y)
C. 2xyz
D. x(2x + y)
E. 2z + 4xy

Answer 1

area of triangle = 1/2 x base x height

Answer 2

Hint: \[A_{\triangle}=\frac{ 1 }{ 2 }bh\]

Answer 3

qwerty@!@@#@#

Answer 4

:D lol

Answer 5

haha

Answer 6

I know that.

Answer 7

But I was thinking about substituting the variables for some numbers.

Answer 8

base =2x+y
height =4z

Answer 9

\[A = \frac{ 1 }{ 2 }(4z)(2x+y)\]

Answer 10

Simplify, all you had to do was plug it in

Answer 11

A=2z(2x+y)

Answer 12

do u wan the solution to that quadrilateral question @Astrophysics ??

Answer 13

YES

Answer 14

ok

Answer 15

lemme solve it XD

Answer 16

So why did it only cancel out the 4 and not the 2. was it to only find half of from only one part (or one number).

Answer 17

ok can u tell what is 2+1??

Answer 18

m not kiddin okay answer this^

Answer 19

it is half base times height
OR
half height times base

half height = 4z/2 = 2z

Answer 20

it works the same as
\[ \frac{1}{2} \cdot 4 \cdot 8 \]
you can group it like this
\[ \left(\frac{1}{2} \cdot 4 \right) \cdot 8\]
only the 4 *1/2 is simplified to 2 and you get 2*8 or 16

if 8 were something complicated like (x+2y) it still be
\[ \left(\frac{1}{2} \cdot 4 \right) \cdot (x+2y)\]

Answer 21

we know that 2+1 = 3

nd if we find out (2+1)/2 then we can't cancel out 2 :)
if u cancell it out then ur left with 1
but we know that 2+1 = 3 nd 3/2 is not 1. so u can't divide like that