Fid the log of 0.0007356
1 answer:
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Answer:
see the explanation
The graph in the attached figure
Step-by-step explanation:
we have
This is a linear equation (the graph is a line)
The y-intercept is the point (0,4) ---> value of y when the value of x is equal to zero)
The x-intercept is the point (-4/3,0) ---> value of x when the value of y is equal to zero)
The slope is m=3
If the slope is changed to m=1/3
we have
This is a linear equation too (the graph is a line)
The y-intercept is the point (0,4) ---> is the same
The x-intercept is the point (-12,0) ---> the x-intercept is different
The slope is m=1/3
The second line becomes a lot less steep
Remember that the slope is the ratio between the rise and the run
In the first line the ratio rise/run is 4/1
In the second line the ratio rise/run is 1/3
see the attached figure to better understand the problem
Answer:
It would take 5.9 years to the nearest tenth of a year
Step-by-step explanation:
The formula of the compound continuously interest is A = P , where
- A is the value of the account in t years
- P is the principal initially invested
- e is the base of a natural logarithm
- r is the rate of interest in decimal
∵ Serenity invested $2,400 in an account
∴ P = 2400
∵ The account paying an interest rate of 3.4%, compounded continuously
∴ r = 3.4% ⇒ divide it by 100 to change it to decimal
∴ r = 3.4 ÷ 100 = 0.034
∵ The value of the account reached to $2,930
∴ A = 2930
→ Substitute these values in the formula above to find t
∵ 2930 = 2400
→ Divide both sides by 2400
∴ =
→ Insert ㏑ in both sides
∴ ㏑() = ㏑(
)
→ Remember ㏑() = n
∴ ㏑() = 0.034t
→ Divide both sides by 0.034 to find t
∴ 5.868637814 = t
→ Round it to the nearest tenth of a year
∴ t = 5.9 years
∴ It would take 5.9 years to the nearest tenth of a year