Gcf is the grouping of all the factors common to both groups
lcm is a group of all the factors that has all the elements of both groups
factor
20x^2y^3=2*2*5*x*x*y*y*y
60x^3y^2=2*2*3*5*x*x*x*y*y
40xy^2=2*2*2*5*x*y*y
the GCF=2*2*5*x*y*y=20xy^2
the LCM=2*2*2*3*5*x*x*x*y*y*y=120x^3y^3
Answer:Angle A is equal to 113
Step-by-step explanation:
If two angles are vertical, that would make them equal to each other.
7x-20=6x-1
Add on to both sides
7x-19=6x
Subtract Seven from both sides
-19=-x
Divide by negative one to get rid of the negative
19=x
Then plug x into the equation 7(19)-20
Answer:
um......I think 2 figure it out....
Step-by-step explanation:
A car has 15 gallons of gas in its tank. The car travels 35 miles per gallon of gas. It uses 1/35 of gas to go 1 mile.
1 imperial foot has approximately 30.48 metric centimeters, and for one present we need 3 ft, or namely 3(30.48) cm, how many can we get from 102 cm? 102 ÷ 3(30.48) ≈ 1.1155, so barely just one present.
Answer:
Step-by-step explanation:
You are being asked to compare the value of a growing infinite geometric series to a fixed constant. Such a series will always eventually have a sum that exceeds any given fixed constant.
__
<h3>a)</h3>
Angelina will get more money from the Choice 1 method of payment. The sequence of payments is a (growing) geometric sequence, so the payments and their sum will eventually exceed the alternative.
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<h3>c)</h3>
For a first term of 1 and a common ratio of 2, the sum of n terms of the geometric series is given by ...
Sn = a1×(r^n -1)/(r -1) . . . . . . . . . . series with first term a1, common ratio r
We want to find n such that ...
Sn ≥ 1,000,000
1 × (2^n -1)/(2 -1) ≥ 1,000,000
2^n ≥ 1,000,001 . . . . add 1
n ≥ log(1,000,001)/log(2) . . . . . take the base-2 logarithm
n ≥ 19.93
The total Angelina receives from Choice 1 will exceed $1,000,000 after 20 days.
