Answer:
= 391/11
Let say number = N
a certain number is increased by 5
= N + 5
,one-half of the result
= (N + 5)/2
Three -fifths of the excess of 61 over the number.
= (3/5)(61 - N)
Equating both
(N + 5)/2 = (3/5)(61 - N)
multiplying by 10 both sides
=> 5(N + 5) = 6(61 - N)
=> 5N + 25 = 366 - 6N
=> 11N = 391
=> N = 391/11
Step-by-step explanation:
<span>1) We are given that PA = PB, so PA ≅ PB by the definition of the radius.
</span>When you draw a perpendicular to a segment AB, you take the compass, point it at A and draw an arc of size AB, then you do the same pointing the compass on B. Point P will be one of the intersections of those two arcs. Therefore PA and PB correspond to the radii of the arcs, which were taken both equal to AB, therefore they are congruent.
2) We know that angles PCA and PCB are right angles by the definition of perpendicular.
Perpendicularity is the relation between two lines that meet at a right angle. Since we know that PC is perpendicular to AB by construction, ∠PCA and ∠PCB are right angles.
3) PC ≅ PC by the reflexive property congruence.
The reflexive property congruence states that any shape is congruent to itself.
4) So, triangle ACP is congruent to triangle BCP by HL, and AC ≅ BC by CPCTC (corresponding parts of congruent triangles are congruent).
CPCTC states that if two triangles are congruent, then all of the corresponding sides and angles are congruent. Since ΔACP ≡ ΔBCP, then the corresponding sides AC and BC are congruent.
5) Since PC is perpendicular to and bisects AB, P is on the perpendicular bisector of AB by the definition of the perpendicular bisector.
<span>The perpendicular bisector of a segment is a line that cuts the segment into two equal parts (bisector) and that forms with the segment a right angle (perpendicular). Any point on the perpendicular bisector has the same distance from the segment's extremities. PC has exactly the characteristics of a perpendicular bisector of AB. </span>
Answer: it could end in 20 different ways.
Step-by-step explanation:
This match could be
3x0 D or F
3x1 D or F
3x2 D or F
For 3x0
We have 2 options: DDD or FFF
2.P₁ = 2.1 = 2
For 3x1
We have 6 options
DFDD FDFF
DDFD FFDF
FDDD DFFF
2.3.1 = 6
For 3x2
we have 12 options:
DFDFD FDFDF
DDFFD FFDDF
DFFDD FDDFF
FFDDD DDFFF
FDFDD DFDFF
FDDFD DFFDF
2.3.2 = 12
In total we could have 20 outcomes.
