Answer:
B. Leg-Acute (LA)
E. Angle-Angle-Side (AAS)
Step-by-step explanation:
Congruence is the relationship between two or more shapes with respect to their common properties.
Comparing the properties of triangles LMN and OPQ, it would be observed that two angles are similar and one side.
So that the congruence theorems or postulates required are:
B. Leg-Acute (LA)
E. Angle-Angle-Side (AAS)
Answer:
x = -2/3; x = 5/2.
Step-by-step explanation:
6x 2 - 11x - 10 = 0
Use the 'ac' method:
6 * -10 = -60. We need 2 numbers whose product is -60 and whose sum is -11.
That would be -15 and +4. So we write:
6x^2 - 15x + 4x - 10 = 0 Factor by grouping:
3x(2x - 5) + 2(2x - 5) = 0
(3x + 2)(2x - 5) = 0
x = -2/3; x = 5/2.
Answer:
A) 7 meters
Step-by-step explanation:
As you know the length of the rectangle you would first multiply the length of 35 by 2, which gives you 70. This is because the perimeter is the sum of all the lengths and that a rectangle has 2 sets of different identical lines. Then you would minus 70 from 84, which gives you 14. This is because you are now finding out the sum of the set of widths. Then you would divide 14 by 2 to find the width, which gives you 7 meters.
1) Multiply 35 by 2.
2) Minus 70 from 84.
3) Divide 14 by 2.
Part A:
To determine the values of the times to which the height of the two cannon balls are the same, we equate the given functions.
H(t) = g(t)
Substitute the expressions for each.
-16t² + 48t + 12 = 10 + 15.2t
Transpose all the terms to the left-hand side of the equation.
-16t² + (48 - 15.2)t + (12 - 10) = 0
Simplifying,
-16t² + 32.8t + 2 = 0
The values of t from the equation are 2.11 seconds and -0.059 seconds
Part B:
In the context of the problem, only 2.11 seconds is acceptable. This is because the second value of t which is equal to -0.059 seconds is not possible since there is no negative value for time.
Answer:
first option
Step-by-step explanation:
Given 2 secants to a circle from an external point, then
The product of the external part and the whole of one secant is equal to the external part and the whole of the other secant, that is
PQ(RP) = PS(TP)