21+4(32-5)=448
I think it’s 448
a+b+c=0
[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]
[a^2+b^2+c^2+2ab+2ac+2bc=0]
[a^2+b^2+c^2=-(2ab+2ac+2bc)]
[a^2+b^2+c^2=-2(ab+ac+bc)] (i)
also
[a=-b-c]
[a^2=-ab-ac] (ii)
[-c=a+b]
[-bc=ab+b^2] (iii)
adding (ii) and (iii) ,we have
[a^2-bc=b^2-ac] (iv)
devide (i) by (iv)
[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
Answer:
d hope it helps make brainlliest ty
Step-by-step explanation:
someone report my question in here
Answer:
<u>Using below system of inequalities</u>
<u>Following the rules </u>
- 1. Finding x- and y - intercepts
- 2. Connecting with dotted line for each as no equal symbol present in any inequality
- 3. Shade respective regions
- 4. Solution is the intersection of the shades regions
- 5. Select any three points in the solution region
<u>Line 1</u>
- y > 2x - 3
- x- intercept: y = 0 ⇒ 0 = 2x - 3 ⇒ 2x = 3 ⇒ x = 1.5
- y - intercept: x = 0 ⇒ y = -3
- Shaded region is above the line (or to the left)
<u>Line 2</u>
- y < x + 1
- x- intercept: y = 0 ⇒ 0 = x + 1 ⇒ x = -1
- y - intercept: x = 0 ⇒ y = 1
- Shaded region is below the line (or to the right)
<u>Selected points are:</u>
Answer:
see explanation
Step-by-step explanation:
Given the recursive formula
= 
with a₁ = 1, then
a₂ =
a₁ =
× 1 = 
a₃ =
a₂ =
×
= 