Answers:
A) SAS
B) ASA
E) LL
============================================================
Explanation:
Let's go through each possible answer choice
A. We can use SAS because we know that AC = XZ (horizontal sides) is one pair of congruent sides, the angle pairing is C = Z, and the other pair of sides is BC = YZ (vertical sides)
B. We can also use ASA. Note how A = X is one pair of angles, AC = XZ is the middle pair of sides, and C = Z is the second pair of angles. The side is between the angles.
C. We can't use SSS as we don't know AB = XY is true or not. There are no tick marks indicating congruence.
D. We can't use LA here because we don't know the altitudes (LA = leg altitude)
E. We can use LL here because we have two right triangles. The two pairs of legs (AC = XZ and BC = YZ) are congruent based on the tickmarks.
F. We can't use HL similar to the reasoning for choice C. We don't know the hypotenuses are congruent or not. If we knew that AB = XY, then we could use HL (hypotenuse leg).
Answer:
5units
Step-by-step explanation:
The triangle inscribed in the circle is a right angled triangle with angle C being 90°.
The diameter of the circle AB is the hypotenuse of the triangle.
Before we can get the radius, we need to get the diameter AB using Pythagoras theorem.
|AB|² = |AC|²+|BC|²
|AB|² = 6²+8²
|AB|² = 36+64
|AB|² = 100
|AB| = √100
|AB| = 10
Diameter of the circle = 10units
Radius = diameter/2 = |AB|/2
Radius = 10/2
Radius = 5units
Answer:
10
Step-by-step explanation:
the sudents were put in 10 equal groups
10x6=60
Answer:
B
Step-by-step explanation:
The solution is the point of intersection of the 2 lines
The lines intersect at (- 3, - 4 ), then
f(x) = g(x) has solution x = - 3
Step-by-step explanation:
We have, Sorin hits a cricket ball straight up into the air. After 1.5 s, the ball is falling straight down with a speed of 6.5 m/s.
We want to find the initial vertical velocity of the ball.
The relation between initial speed, final speed, acceleration and time is given by using first equation of kinematics. Let v₀ is the initial speed. So,
Here, v = -6.5 m/s, t = 1.5 s and a = -g
So, the initial velocity of the ball is 8.2 m/s and the formula used is (1).
