It's 2/5 in decimal form it is 0.4.
1: yes AAS
2: yes SSS reflective
3: yes ASA
4: yes HL reflective
5: no SSA vertical
6: yes SAS vertical
Not sure on vocabulary for 1 and 3 sorry:(
Answer:
Individuals end to continue paying the premiums of the automobile insurance as a habit. However, serious thoughts and putting in element of strategizing helps to reduce the premium in most cases. At times, there is a sudden like on the part of the insurer even for a flawless driver.
A good look up and research of the insurance websites can be of real help in comparing whether a better deal is offered by the other insurance companies, or whether a certain change in the policy or small adjustments of the term would give benefit to the customer.
In case a speeding ticket is received, or an accident is mentioned in the driving history, it is maintained there in for a period of three to five years. Thus, the premium increases substantially. A change of insurer is advised in such situations, where a major search for an insurer, who does not pay that much importance to these details, is to be carried on.
Again, having a teenager driver in the family calls for a caution as the insurance premium increases drastically in such occasion. Having clean driving record of the parents, or kids commuting to far away schools without cars help in such situation.
You can set up a system of equations for this problem. x= number of coach tickets and y = number of first class tickets.
$210x + $1200y = $10,230 (cost of coach ticket plus cost of first class tickets is total budget)
x + y = 11 (number of coach tickets plus number of first class tickets is total number of people)
Solve the second equation for y to get y = 11 - x, then plug that into the first equation and solve for x:
$210x + $1200(11 - x) = $10,230
$210x + $13,200 - $1200x = $10,230
-$990x + $13,200 = $10,230
-$990x = $2,970
x = 3
Sarah bought x = 3 coach tickets. Plug that into the second equation and solve for y:
3 + y = 11
y = 8
Sarah bought y = 8 first class tickets.
An = a1 + (n - 1)(d)
Where a1 is the first term and d is the common difference.
First find d, the common difference.
24, ____, 32
a3 a4 a5
Subtract 32-24 = 8
Subtract a5 - a3 = 2
Divide 8/2 = 4
d = 4
Use d and one of the values they give us to find a1.
a3 = 24
24 = a1 + (3 - 1)(4)
24 = a1 + 2(4)
24 = a1 + 8
Subtract 8 from both sides
16 = a1
an = 16 + (n - 1)(4)
Can also be written
an = 16 + 4n - 4
an = 4n + 12
