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Answer:
3x+3a-9e+8
Step-by-step explanation:
(4x+4x-5x)+3a-9e+8
8x-5x+3a-9e+8
3x+3a-9e+8
Answer:
HL
Step-by-step explanation:
The things that are congruent between the 2 triangles are the 90° angles and the hypotenuses and legs. HA and LA only work with the acute angles, not right angles, and AA isn't a way of proving things to be congruent, meaning that HL is the answer
Answer:
The value of the side PS is 26 approx.
Step-by-step explanation:
In this question we have two right triangles. Triangle PQR and Triangle PQS.
Where S is some point on the line segment QR.
Given:
PR = 20
SR = 11
QS = 5
We know that QR = QS + SR
QR = 11 + 5
QR = 16
Now triangle PQR has one unknown side PQ which in its base.
Finding PQ:
Using Pythagoras theorem for the right angled triangle PQR.
PR² = PQ² + QR²
PQ = √(PR² - QR²)
PQ = √(20²+16²)
PQ = √656
PQ = 4√41
Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.
Finding PS:
Using Pythagoras theorem, we have:
PS² = PQ² + QS²
PS² = 656 + 25
PS² = 681
PS = 26.09
PS = 26
Answer:
H=9;8)
B=(5;4) (ball)
R=(7;0) (hit point)
B'=symetric of B axis perpendicular of x in R
B'=(7+(7-5);4)=(9;4)
Equation BR: y-4=(0-4)/(7-5)(x-5)==>y=-2x+14
Equation RB': y-4=(4-0)/(9-7)(x-9)==>y=2x-14
Is H a point of RB'? y=2x-14 : 8=? 2*9-14==>8=?4 No!
you will not make your putt
Step-by-step explanation:
