Is there a piece of paper that comes with the questions? Or do I just pick one? If there is a piece of paper attached, please add it to your question. Thank you!
Answer:
Step-by-step explanation:
To prove Δ ABC similar to ΔDBE we can consider
Segments AC and DE are parallel.
⇒ DE intersects AB and BC in same ratio.
AB is a transversal line passing AC and DE.
⇒∠BAC=∠BDE [corresponding angles]
Angle B is congruent to itself due to the reflexive property.
All of them are telling a relation of parts of ΔABC to ΔDBE.
The only option which is not used to prove that ΔABC is similar to ΔDBE is the first option ,"The sum of angles A and B are supplementary to angle C".
Answer:
d = 14/5
Step-by-step explanation:
The point (-4,2) means that;
At x = -4, y = 2
Now general form of a linear equation is;
Ax + By + C = 0
We are given;
4y = 3x + 6
Rearranging to the form of a linear equation gives;
3x - 4y + 6 = 0
Thus, A = 3, B = -4 and C = 6
Thus, at point (-4,2), distance between them is;
d = (3(-4) - 4(2) + 6)/√(3² + (-4)²)
d = -14/5
We will take the absolute value.
Thus; d = 14/5
Answer:
C. -8/3
Step-by-step explanation:
3t + 4 = -4
3t = -4 - 4
3t = -8
t = -8/3
Answer:
Step-by-step explanation:
Just use the distributive property.
(4y^2-3)(4y+3)
= 4y^2(4y+3) - 3(4y+3)
= 16y^3 + 12y^2 - 12y - 9