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Murrr4er [49]
3 years ago
12

Use the line graph to solve.

Mathematics
1 answer:
Ivahew [28]3 years ago
5 0
0.10 m of difference <span>in water levels between Week 2 and Week 4</span>
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Pls help with step by step :)
Talja [164]

Answer:

Step-by-step explanation:

Hope this helps u !!

6 0
3 years ago
Find the slope of this line. *
Elden [556K]

Answer:

1

Step-by-step explanation:

Slope = y2 - y1 / x2 - x1 so, (0 - (-2)) / (1 - (-1)) == 2/2 = 1

3 0
3 years ago
Read 2 more answers
3) An $1,000 investment is made in a trust fund at an annual percentage rate of 12%,
Elina [12.6K]

We need to find the interest rate in order to solve this problem:

.12/12 = (.01)

The number of years is unknown. T represents time in the following formula:

1000 (initial investment)

2000 (balance after t years from investment)

$1000 (1 + .01)^{12t}=$2000

In order to isolate the exponential term, both sides must be divided by 1000.

(1+.01)^{12t}=2

By using natural logarithm:

Ln(1+.01)^{12t} =Ln2\\(12t)Ln(1+.01)=Ln2\\

By dividing both sides of the equation by (12)Ln(1+.01)

t=\frac{Ln(2)}{(12)Ln(1+.01)} =5.805yrs

5 0
3 years ago
Devise the exponential growth function that fits the given​ data, then answer the accompanying question. Be sure to identify the
NemiM [27]

Answer:

The cost in 2015 will be $283.45.

What is the reference point (t=0)?

d. The year 2005.

Step-by-step explanation:

The exponential growth model for the cost is the following:

C(t) = C_{0}e^{rt}

In which C(t) is the cost after t years, C_{0} is the initial cost and r is the decimal inflation rate

In this problem, we have that:

C_{0} = 190, r = 0.04

What will it cost in 2015 assuming the rate of inflation remains​ constant?

2015 is 10 years after 2005, so this is C(10).

C(t) = C_{0}e^{rt}

C(10) = 190*e^{0.04*10} = 283.45

The cost in 2015 will be $283.45.

The reference point is the year in which t = 0, so the year 2005.

7 0
3 years ago
Read 2 more answers
R+12
Mice21 [21]
All of these mean r + 12
6 0
4 years ago
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