Factoring out the GCF of the polynomial 2x^6+2x^5 will give
1 answer:
You might be interested in
See the attached figure.
========================
AB = 10 , FD = 3
∵ D is the midpoint of AB, and F is the mid point of CB
∴ FD // AC , FD = 0.5 AC
∵ Δ ABC is a right triangle at C
∴ FD ⊥ BC
∴ BD = 0.5 AB = 5
∴ in Δ FDB ⇒⇒ BF² = BD² - FD² = 5² - 3² = 16
∴ BF = √16 = 4
∵ F is the mid point of CB
∴ CF = BF = 4 , and CB = 2 BF = 2*4 = 8
∵ D is the midpoint of AB, and E is the mid point of AC
∴ DE // CB , and DE = 0.5 CB = 0.5 * 8 = 4
∴ T<span>he length of line ED is 4
</span>
========================
AB = 10 , FD = 3
∵ D is the midpoint of AB, and F is the mid point of CB
∴ FD // AC , FD = 0.5 AC
∵ Δ ABC is a right triangle at C
∴ FD ⊥ BC
∴ BD = 0.5 AB = 5
∴ in Δ FDB ⇒⇒ BF² = BD² - FD² = 5² - 3² = 16
∴ BF = √16 = 4
∵ F is the mid point of CB
∴ CF = BF = 4 , and CB = 2 BF = 2*4 = 8
∵ D is the midpoint of AB, and E is the mid point of AC
∴ DE // CB , and DE = 0.5 CB = 0.5 * 8 = 4
∴ T<span>he length of line ED is 4
</span>