Here it is given that AB || CD
< EIA = <GJB
Now
∠EIA ≅ ∠IKC and ∠GJB is ≅ ∠ JLD (Corresponding angles)
∠EIA ≅ ∠GJB then ∠IKC ≅ ∠ JLD (Substitution Property of Congruency)
∠IKL + ∠IKC 180° and ∠DLH + ∠JLD =180° (Linear Pair Theorem)
So
m∠IKL + m∠IKC = 180° ....(1)
But ∠IKC ≅ ∠JLD
m∠IKC = m∠JLD (SUBTRACTION PROPERTY OF CONGRUENCY)
So we have
m∠IKL + m∠JLD = 180°
∠IKL and ∠JLD are supplementary angles.
But ∠DLH and ∠JLD are supplementary angles.
∠IKL ≅ ∠DLH (CONGRUENT SUPPLEMENTS THEOREM)
Answer:
SOHCAHTOA.
Step-by-step explanation:
acronym is an abbreviation formed from the initial letters of other words and pronounced as a word
The acronym for sin cosine and tangent is
SOHCAHTOA
Sine =Opposite over Hypotenuse
Cosine= Adjacent over Hypotenuse
Tangent= Opposite over Adjacent
Y=-4x - 24
I hope this helps you :)
Answer:
recursive: f(0) = 7; f(n) = f(n-1) -8
explicit: f(n) = 7 -8n
Step-by-step explanation:
The sequence is an arithmetic sequence with first term 7 and common difference -8. Since you're numbering the terms starting with n=0, the generic case will be ...
recursive: f(0) = first term; f(n) = f(n-1) + common difference
explicit: f(n) = first term + n·(common difference)
To get the answer above, fill in the first term and common difference values.
You first find the average of the first two test scores. When you find the average, you add the two numbers together and then divide by two which gives you 89. You then plug in different scores above the number 89 and find the average of those (89 and x) to see which gives you an average of 92. It’s mostly a guess and check type of problem.
