Answer:
Given: Quadrilateral P QR S is a rectangle.
To prove :PR= Q S
Construction : Join PR and Q S.
Proof: In Rectangle PQRS, and
→ taking two triangles PSR and Δ QRS
1. PS = Q R
2. ∠ PS R = ∠ Q RS [Each being 90°]
3. S R is common.
→ ΔP SR ≅ Δ Q RS → [Side-Angle-Side Congruency]
∴ PR =Q S [ corresponding part of congruent triangles ]
Hence proved.
If y=1, then y^5 is 1*1*1*1*1=1.
If cente ris (h,k) and radius is r then
equatn is (x-h)^2+(y-k)^2=r^2
so
center (5,4)
r=2
h=5
k=4
(x-5)^2+(y-4)^2=2^2
(x-5)^2+(y-4)^2=4
first option
-5.55 - 8.55c + 4.35c
-8.55c + 4.35c = -4.2c
-5.55 - 4.2c is the answer.
Answer cosine- cos- can =adj/hyp
Step-by-step explanation:
Ummm try this