Answer:
Step-by-step explanation:
x = 8 ; y = 5/3

![\dfrac{1}{4}[x(2y+3z)] =\dfrac{1}{4}[8*(2*\dfrac{5}{3}+\dfrac{5}{3})]\\\\=\dfrac{1}{4}(8*(\dfrac{10}{3}+\dfrac{5}{3})]\\\\=\dfrac{1}{4}(8*\dfrac{15}{3})\\\\=\dfrac{1}{4}*8*5\\\\=2*5\\\\=10](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B4%7D%5Bx%282y%2B3z%29%5D%20%3D%5Cdfrac%7B1%7D%7B4%7D%5B8%2A%282%2A%5Cdfrac%7B5%7D%7B3%7D%2B%5Cdfrac%7B5%7D%7B3%7D%29%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B4%7D%288%2A%28%5Cdfrac%7B10%7D%7B3%7D%2B%5Cdfrac%7B5%7D%7B3%7D%29%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B4%7D%288%2A%5Cdfrac%7B15%7D%7B3%7D%29%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B4%7D%2A8%2A5%5C%5C%5C%5C%3D2%2A5%5C%5C%5C%5C%3D10)
It’s a cool shape. Reminds me a flash.
Answer:
Step-by-step explanation:
Given
Two sides of triangle of sides 5 ft and 7 ft
and angle between them is increasing at a rate of 0.9 radians per second
let
is the angle between them thus
Area of triangle when two sides and angle between them is given


Differentiate w.r.t time

at 


Answer:
In quadrilateral ABCD we have
AC = AD
and AB being the bisector of ∠A.
Now, in ΔABC and ΔABD,
AC = AD
[Given]
AB = AB
[Common]
∠CAB = ∠DAB [∴ AB bisects ∠CAD]
∴ Using SAS criteria, we have
ΔABC ≌ ΔABD.
∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.
∴ BC = BD.