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.
Given:
Polynomials
To find:
Monomial of 2nd degree with leading coefficient 3
Solution:
Monomial is an algebraic expression with only one term.
Option A:
It is not a monomial because it have 2 terms.
It is not true.
Option B:
It is not a monomial because it have 2 terms.
It is not true.
Option C:
It have one term only. So, it is a monomial.
Degree means highest power. So degree = 2
Leading coefficient means the value before variable.
Leading coefficient = 3
It is true.
Option D:
It have one term only. So, it is a monomial.
Degree means highest power. So degree = 3
It is not true.
Therefore is a monomial of 2nd degree with a leading coefficient of 3.
Domain:
Range:
We are given the set of ordered pairs: . The Domain is just the set containing all the
values of each ordered pair of
. Recall that in the ordered pair
,
is the
value. In set
, we can see that the
values are
.
For the Range, it is just the set containing all the values of each ordered pair of
. Recall that in the ordered pair
,
is the
value. In set
, we can see that the
values are
.