Formula of the circle
(x-a)² +(y-b)² = r²,
where r is a radius, (a,b) - coordinates of the center.
So,
x² + y² = 49
(x-0)² + (y-0)² = 49.
Center of this circle has coordinates (0,0).
x² + y² = 7²
r = 7
Answer is C.7.
Answer:
Step-by-step explanation:
QM is the angle bisector of ∠LMP
∠LMQ = ∠QMP
QM is the angle bisector of ∠PQL
∠PQM = ∠MQL
MQ = QM as common
By ASA, triangle MQP ≅ MQL
LM = PM and LQ = PQ as they are same side of congruent triangles
Triangle LPQ and LPM are isosceles
By angle bisector theorem, LP is perpendicular to MQ
By properties of rhombus, the two diagonals are perpendicular proves that LMPQ is a rhombus.
LM ≅ PQ
Answer:
[-20.25;+∞).
Step-by-step explanation:
1) according to the properties of the given parabola the minimum is:
2) y₀=-20.25 means, the value -20.25 is minimum of the range, the maximum →+∞;
3) finally, the range y≥-20.25.
We have that
<span>tan(theta)sin(theta)+cos(theta)=sec(theta)
</span><span>[sin(theta)/cos(theta)] sin(theta)+cos(theta)=sec(theta)
</span>[sin²<span>(theta)/cos(theta)]+cos(theta)=sec(theta)
</span><span>the next step in this proof
is </span>write cos(theta)=cos²<span>(theta)/cos(theta) to find a common denominator
so
</span>[sin²(theta)/cos(theta)]+[cos²(theta)/cos(theta)]=sec(theta)<span>
</span>{[sin²(theta)+cos²(theta)]/cos(theta)}=sec(theta)<span>
remember that
</span>sin²(theta)+cos²(theta)=1
{[sin²(theta)+cos²(theta)]/cos(theta)}------------> 1/cos(theta)
and
1/cos(theta)=sec(theta)-------------> is ok
the answer is the option <span>B.)
He should write cos(theta)=cos^2(theta)/cos(theta) to find a common denominator.</span>
