Answer:
a. S = 3n + 2
b. There while be 62 squares.
Step-by-step explanation:
We know the first term of this sequence is 5. To figure out the equation, subtract the following term from the previous. Do you see a common difference?
8 - 5 = 3
11 - 8 = 3
14 - 11 = 3
We're seeing a constant difference of 3 (which makes this an arithmetic sequence), but the first term is 5. That mean something is being added to make the first term 5. Subtract 3 from 5 to get 2. This means 2 is being added to every multiple of 3, which leads us to the equation: S = 3n + 2.
To find the 20th term of this sequence, substitute n for 20 and do the operations.
S = 3(20) + 2
<em>Multiply 3 by 20, then add 2.</em>
S = 62
The 20th term will have 62 squares.
Problem Example :
Given: HF || JK; HG ≅ JG
Prove: FHG ≅ KJG
Answer: A. ∠FGH ≅∠KGJ because vertical angles are congruent.
Answer:
0.2 centimeters
Step-by-step explanation:
Strategy 1:
Make the equation: 5=a•s, where s is the scale factor and "a" is the number of centimeters which 5 meters
First, we need to find the scale factor. 50 meters would be s times greater than 2 centimeters. Find s by first converting meters into centimeters
50m=5000cm. 5000cm=2cm*s, so s = 2500
5m=a*2500, a=0.002 meters, and 0.002 meters is equal to 0.2 centimeters.
Strategy 2:
the ratio of the scale drawing to real life will always stay the same, so
2 cm / 50 meters = x / 5 meters, and cross multiply to get
x cm* 50 m = 2 cm * 5 m, so
x cm * 10 = 2 cm
x cm = 2/10 cm
x cm = 0.2 cm
Strategy 3:
Notice that 5 meters is 10 times smaller than 50 meters, so on the scale drawing, we are looking for a number 10 times smaller than 2 centimeters, so 2/10=0.2 cm
A short-cut to accurately evaluate the given expression above is using a scientific calculator where one can include integrals and evaluate using limits. In this case, using a calculator, the answer is equal to 0.2679. One can verify this by integrating truly letting 1-x^2 as u and use its du to be substituted in the numerator
There is a 12 degrees difference in percentage for the temperature.
