Evaluate: 2(1/4) + 2/5 ÷ (-1/2) - 1/4
⇛9/4 + 2/5 ÷ (-1/2) - 1/4
Take the LCM of the 4 and 5 is 20.
Again, 2 and 4 is 4.
⇛[(54+8)/20] ÷ [(-2 - 1)/4]
⇛62/20 ÷ (-3/4)
⇛62/20 × 4/-3
⇛62/{5*(-3)}
⇛62/-15.
The answer is (3x - 4) (x + 3)
For E2020 users, the correct answer is A (No,<span>because </span>P<span>(F ∩ T) ≠ </span>P<span>(F) • </span>P<span>(T).)</span>
Answer:
first three
Step-by-step explanation:
go directly to the x- value ed2020
Given : In Right triangle ABC, AC=6 cm, BC=8 cm.Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5.
To find : Area (ΔMNC)
Solution: In Δ ABC, right angled at C,
AC= 6 cm, BC= 8 cm
Using pythagoras theorem
AB² =AC²+ BC²
=6²+8²
= 36 + 64
→AB² =100
→AB² =10²
→AB =10
Also, AM:MN:NB=1:2.5:1.5
Then AM, MN, NB are k, 2.5 k, 1.5 k.
→2.5 k + k+1.5 k= 10
→ 5 k =10
Dividing both sides by 2, we get
→ k =2
MN=2.5×2=5 cm, NB=1.5×2=3 cm, AM=2 cm
As Δ ACB and ΔMNC are similar by SAS.
So when triangles are similar , their sides are proportional and ratio of their areas is equal to square of their corresponding sides.
![\frac{Ar(ACB)}{Ar(MNC)}=[\frac{10}{5}]^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BAr%28ACB%29%7D%7BAr%28MNC%29%7D%3D%5B%5Cfrac%7B10%7D%7B5%7D%5D%5E%7B2%7D)

But Area (ΔACB)=1/2×6×8= 24 cm²[ACB is a right angled triangle]

→ Area(ΔMNC)=24÷4
→Area(ΔMNC)=6 cm²