If x=3 is a solution of the equation 2x²-5x+k=0 find the value of k
2 answers:
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Answer:
Chandra's Average Monthly Expenses are;
1) For Both Jan and Feb are $269
2) For the remaining 10 months each are $244
Step-by-step explanation:
From Chandra's matrix all monthly expenses are all same that is,
Cell phone $71, Rent $1,025, Gym $75, Internet $25, Auto insurance $425, Gas $ 120, Food $145. which are all the expenses carried out every month for 12 months.
That means Chandra carries out a total of 7 expenses for the month of January and February while she carried a total of 6 expenses for the remaining 10 month which was noted from the matrix that Auto insurance was carried out only in the month of January and February.
Therefore, you start by adding up each month total expenses, which are ;
January = $71 + $1,025 + $75 + $25 + $425 + $120 + $145 = $1886
February = $71 + $1,025 + $75 + $25 + $425 + $120 + $145 = $1886
March = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
April = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
May = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
June = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
July= $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
August = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
September = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
October = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
November = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
December = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
Therefore Chandra's Average Monthly Expenses are:
1) January & February = =
= 269.4286 to the nearest cent
≅ $269
2) For the remaining 10 month are = =
= 243.5 to the nearest cent
≅ $244
Answer:
Step-by-step explanation:
Having drawn the line, Kendall must verify that the point P belongs to the line y = 2x-1 and then calculate the distance between A-P and verify if it is the closest to A or there is another one of the line
Having the point P(3,5) substitue x to verify y
y=2*(3)-1=6-1=5 (3,5)
Now if the angle formed by A and P is 90º it means that it is the closest point, otherwise that point must be found
and we found the distance PQ and QA
; ,
be the APQ triangle we must find <APQ through the cosine law (graph 2).
