<h2>(1)</h2><h2> =(a+b)(3c-d)</h2><h2> =a(3c-d)+b(3c-d)</h2><h2> =3ac-ad+3bc-bd</h2>
<h2>(2)</h2><h2> =(a-b)(c+2d)</h2><h2> =a(c+2d)-b(c+2d)</h2><h2> =ac+2ad-bc-2bd</h2>
<h2>(3)</h2><h2> =(a-b)(c-2d)</h2><h2> =a(c-2d)-b(c-2d)</h2><h2> =ac-2ad-bc+2bd</h2>
<h2>(4)</h2><h2> =(2a+b)(c-3d)</h2><h2> =2a(c-3d)+b(c-3d)</h2><h2> =2ac-6ad+bc-3bd</h2>
Answer: Simple and best practice solution for 4y^2+8y+1=0 equation. ... Begin completing the square. ... Square it (1) and add it to both sides.
Step-by-step explanation: Complete the square for x2+2x x 2 + 2 x.
I would say 19.
If you take out the 6s and just look at the rest you have 4, 9, 14.
It is adding 5 each time and just have a 6 between each one.
So logically I would say 19.
The main rules that we use here are :
i)
for nonnegative values a and b.
ii)
.
Thus, first 'decompose' the numbers in the radicals into prime factors:
.
By rule (i) we write:
.
We can collect these terms as follows:
, and by rule (ii) we have:
Answer:
.
5x^3/10x^2
*Simplify 5/10 to 1/2
*simplify the exponent so that x^1 is on top
1x/2
FINAL ANSWER:
x/2
