Answer:
Proven
Step-by-step explanation:
Let ABC be a triangle and D, E and F are midpoints of BC, CA and AB.
The sum of two sides of a triangle is greater than twice the median bisecting the third side,
Hence in ΔABD, AD is a median
⇒ AB + AC > 2(AD)
Hence, we get
BC + AC > 2 (CF
)
BC + AB > 2 (BE)
On adding the above inequalities, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2 (AD) + 2 (CD) + 2 (BE
)
2(AB + BC + AC) > 2(AD + BE + CF)
AB + BC + AC > AD + BE + CF - Proven
* Sorry I don't have any attachments, use your imagination!
Answer:
b. (3,0)
e. (5,-6)
f. (2,1)
Are solutions to the system of inequalities
Step-by-step explanation:
b. (3,0)
0 ≤ -2(3) + 6
0 ≤ 0 is true
3 > 1 is true
e. (5,-6)
-6 ≤ -2(5)+6
-6 ≤ -4 is true
5 > 1 is true
f. (2,1)
1 ≤ -2(2)+6
1 ≤ 2 is true
2 > 1 is true
Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)
first factor into (x-r1)(x-r2)... form
p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16)
group the like ones
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
multiplicity is how many times the root repeats in the function
for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2
so
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1
(x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real
baseically
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1
Answer:
No
Step-by-step explanation:
3x-5= 4+2x
Or, x=4+5
x=9
