Step-by-step explanation:
20.
In each proof, start by looking at what you're trying to prove. We want to prove that two triangles are congruent. To do that we use one of the following: SSS, SAS, ASA, or AAS.
To decide which one to use, look at the information given. We're given two pairs of congruent sides, so we can narrow the strategy down to either SSS or SAS. We aren't told anything about the third pair of sides, but we <em>can</em> see that ∠JNK and ∠MNL are vertical angles. We'll use this to show the triangles are congruent by SAS.
1. JN ≅ MN, Given
2. ∠JNK ≅ ∠MNL, Vertical angles
3. NK ≅ NL, Given
4. ΔJNK ≅ ΔMNL, SAS
21.
Repeat the same steps as 20. Again, we're trying to prove two triangles are congruent, so we have 4 strategies to choose from. Just like before, we're given two pairs of congruent sides, so we'll use either SSS or SAS. And again, we aren't told anything about the third pair of sides, but we can see that both triangles are right triangles. So we'll use SAS again.
1. MN ≅ PQ, Given
2. ∠LMN ≅ ∠NQP, Right angles are congruent
3. LM ≅ NQ, Given
4. ΔNML ≅ ΔPQN, SAS
Answer:
Ashley would have to pay an additional $608.52 in interest when the length of the loan changes from 3 years to 5 years.
Step-by-step explanation:
Although the problem gives us the initial amount of the balance ($4000) and the interest rate (13%), we are really only concerned with the total amount Ashley will make in payments in three years versus five years. If she pays $134.78 for three years, that is 36 months, or $134.78 times 36 which is equal to $4852.08. However, if she decides to take longer to pay it and instead pays $91.01 for five years, or 60 months, then her total amount paid would be $5460.60. To find the difference between these two, simply subtract $5460.60-$4852.08 to get $608.52.
8/12 = 12/16
step by step equation
Answer:
Sequence One
Step-by-step explanation:
S = a₁ / (1 − r) is the infinite sum of a geometric series where |r| < 1.
Sequence One: r = 1/5
Sequence Two: r = 2
Sequence Three: not geometric
Sequence Four: r = -3/2
Only Sequence One has |r| < 1.
Count the number of multiples of 3, 4, and 12 in the range 1-2005:
⌊2005/3⌋ ≈ ⌊668.333⌋ = 668
⌊2005/4⌋ = ⌊501.25⌋ = 501
⌊2005/12⌋ ≈ ⌊167.083⌋ = 167
(⌊<em>x</em>⌋ means the "floor" of <em>x</em>, i.e. the largest integer smaller than <em>x</em>, so ⌊<em>a</em>/<em>b</em>⌋ is what you get when you divide <em>a</em> by <em>b</em> and ignore the remainder)
Then using the inclusion/exclusion principle, there are
668 + 501 - 2•167 = 835
numbers that are multiples of 3 or 4 but not 12. We subtract the number multiples of 12 twice because the sets of multiples of 3 and 4 both contain multiples of 12. Subtracting once removes the multiples of 3 <em>and</em> 4 that occur twice. Subtracting again removes them altogether.
