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3 years ago

What value must k take in order for the following expression to be greater than zero?

Mathematics

1 answer:

svetlana [45]3 years ago

8 0

Basically anything greater than zero...k>0

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In 2000, the population of Big Springs was 13 thousand. Use the given doubling

trasher [3.6K]

Answer:

The answer is "26179.4".

Step-by-step explanation:

Assume year 2000 as t, that is t =0.

Formula:

$A= A_0e^{rt}$

Where,

$A_0 = \ initial \ pop \\\\r= \ rate \ in \ decimal \\\\t= \ time \ in \ year$

for doubling time,

$r = \frac{log (2)}{t} \\$

$r = \frac{\log (2)}{ 40} \\\\r= \frac{0.301}{40}\\\\r= 0.007$

Given value:

$A = A_0e^{rt} \\\\$

$A_0 = 13000$

$t= 40 \ years$

when year is 2000, t=0 so, year is 2100 year as t = 100.

$A = 13000 \times e^{et}\\\\A = 13000 \times e^{e \times t}\\\\A = 13000 \times e^{0.007 \times 100}\\\\A = 13000 \times e^{0.7}\\\\A= 13000\times 2.0138\\\\A = 26179.4$

7 0

3 years ago

Find the product of (3x²y²) (4xy³)²

Elena-2011 [213]

First, I would solve the second parenthesis.

(4xy^3)^2

Distribute

(4^2 x^2 y^6)

4 x 4 = 16

Now, combine like terms

3x^2 x 4x^2 = 12x^4

y^2 x y^6 = y^12

So, the answer would be 12x^4 y^12

Hmm actually I'm not sure. I did this about two years ago so I don't really remember sorry if this is really wrong

5 0

3 years ago

Read 2 more answers

If the mass of a material is 44 grams and the volume of the material is 6 cm^3, what would the density of the material be?

Svetlanka [38]

Density is mass per unit volume.

For the given material that has a mass of 44 grams occupying a volume of 6 cm^3, we divide the mass by the volume, i.e.

density of material = 44 grams / 6 cm^3 = 44/6 grams/cm^3 = 7.3 g/cm^3 (to two significant figures)

4 0

3 years ago

Angelina paid $6.60 for a dozen cupcakes for a birthday party. What is the unit cost for each cupcake? *

S_A_V [24]

Answer:$.55

Step-by-step explanation:

An important thing to know is that a dozen is equal to 12.

12x= 6.60

/12 /12

x=.55

3 0

3 years ago

Read 2 more answers

The graph below represents a population

Nataliya [291]

Answer:

Average rate of change for the function for the interval (6, 12] is 500 people per year.

Option A is correct.

Step-by-step explanation:

We need to find the average rate of change for the function for the interval

(6, 12]

The formula used to calculate Average rate of change is:

$Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}$

We are given a=6 and b=12

Looking at the graph we can see that when x=6 y= 3000 so, f(a)=3000

and when x=12, y=6000 so, f(b)=6000

Putting values in formula and finding Average rate of change:

$Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}\\Average \ rate \ of \ change=\frac{6000-3000}{12-6}\\Average \ rate \ of \ change=\frac{3000}{6}\\Average \ rate \ of \ change=500$

So, average rate of change for the function for the interval (6, 12] is 500 people per year.

Option A is correct.

6 0

3 years ago