What is the formula for Simple Interest? ...What does each letter mean?
1 answer:
You might be interested in
To get the answer, you need to think of it like a see-saw.
Both sides must equal the same relative to 33.5. For example, 32 is 1.5 away from 33.5 so you could say you have 1.5*4 (because theres 4 tubes with 32 sweets). If you were to do this for the whole table, it would look something like this:
This simplifies to:
This method can sometimes be inefficient depending on how many values but is probably the simplest way to solve this problem.
hope this helps :)
(sorry the other way is kinda hard to explain possibly ask your teacher tomorrow)
Both sides must equal the same relative to 33.5. For example, 32 is 1.5 away from 33.5 so you could say you have 1.5*4 (because theres 4 tubes with 32 sweets). If you were to do this for the whole table, it would look something like this:
This simplifies to:
This method can sometimes be inefficient depending on how many values but is probably the simplest way to solve this problem.
hope this helps :)
(sorry the other way is kinda hard to explain possibly ask your teacher tomorrow)
Answer:
The correct options are;
EFGH has 4 congruent sides
Diagonal FH bisects angles EFG and EHG
Angle FEH is congruent to angle FGH
Step-by-step explanation:
1) Given that for a reflection, we have;
The distance of the reflected preimage from the line of reflection = The distance of the reflected image from the line of reflection
Therefore;
The distance of the point E from the line HF = The distance of the point G from the line HF
Also the reflection of an preimage (x, y) about the x-axis, gives an image (x, -y)
We can show that from the length of a line given by the equation, that the length EH ≅ GH and EF ≅ GF
Therefore since we are given that EH = EF, we have;
EH = GH = GF = EF by the definition of congruency, which gives 4 congruent sides
2) Given that EH = GH = GF = EF and HF = FH by reflective property, we have;
ΔEHF ≅ ΔGHF
∴ ∠GHF ≅ ∠EHF by Congruent Parts of Congruent Triangles are Congruent
Similarly, ∠GFH ≅ ∠EFH
Therefore, ∠GFH = ∠EFH and ∠GHF = ∠EHF
Therefore, diagonal FH bisects angles EFG and EHG
3) Given that ΔEHF ≅ ΔGHF, we have;
Angle FEH is congruent to angle FGH, by Congruent Parts of Congruent Triangles are Congruent