step 4: x^2+4=10 |work square x and 2|
step 5: x^2=6 |work subtract 4 from both sides|
step 6: X = 2.44 |work find the root of 6|
Answer:
3x^2-7x-20
Step-by-step explanation:
Given: An Isosceles trapezoid EFGH in which EF =GH
To prove: ΔFHE ≅ ΔGEH
Proof: In Isosceles trapezoid EFGH, Considering two triangles ΔFHE and ΔGEH
1. FE ≅ G H → [ Given]
2. ∠H = ∠E
→ Draw GM⊥HE and FN ⊥EH, and In Δ GMH and ΔFNE,
GH=FE [Given]
∠M+∠N=180° so GM║FN and GF║EH, So GFMN is a rectangle.]
∴ GM =FN [opposite sides of rectangle]
∠GMH = ∠FNE [ Each being 90°]
Δ GMH ≅ ΔFNE [ Right hand side congruency]
→∠H =∠E [CPCT]
→ Side EH is common i.e EH ≅ EH .
→ΔFHE ≅ ΔGEH. [SAS]
U(x)=-2x²+3
v(x)=1/x →→ range of (uоv)(x, BUT range (uov)(x = u(v(x)), then :
u(v(x)) = -2(1/x²) + 3 →→u(v(x)) = -2/x² + 3
The range of -2/x² + 3 = {x∈R/x≠0}
Answer:
6600 yd²
Step-by-step explanation:
The area of the scale drawing is ...
(5.5 in)(3 in) = 16.5 in²
Each square inch represents an area of the field that is ...
(20 yd)×(20 yd) = 400 yd² . . . . per square inch of drawing
Then the scale drawing represents a field with an area of ...
(16.5 in²)×(400 yd²/in²) = 6600 yd²