Write a whole number and a fraction greater than 1 to name the part filled .

Evaluate the expression below for x= 5.<br><br>3x+ 2 · |x-81 - 12

Natasha2012 [34]

Hopefully that’s a one

4 0

3 years ago

Please solve this geometry equation with the work ! for 40 points and brainliest

topjm [15]

Answer:

D. 16 times.

Step-by-step explanation:

Volume of a cylinder = π r^2 h

π r^2 h = 432

Now if the diameter is reduced by 1/4 then the radius is reduced by a fourth also.

Let the new height be H, then:

432 = π (r/4)^2 H

so π r^2 h = π(r/4)^2 H

r^2 h = r^2 H / 16

h = H/16

H = 16h.

5 0

3 years ago

Consider the following simplified algebraic expression. Which could be the original expression?

dem82 [27]

You have to simplify each equation. You can do that by adding similar values to each other:

(the first equation)

8x²y + 7x² - 5y² + 3x - 2 + 2x²y - x² - 2y² + 3x - 4

To make it easier, arrange the values around so that similar ones are beside each other:

8x²y + 2x²y + 7x² - x² - 5y² - 2y² + 3x + 3x - 2 - 4

then add the similar values together:

8x²y + 2x²y = 10x²y

7x² - x² = 6x²

- 5y² - 2y² = -7y²

3x + 3x = 6x

(- 2)+ (- 4) = -6

and if you put them together:

10x²y + 6x² - 7y² + 6x - 6

This isn't the answer to the equation, because it's different. So repeat the process with all the other equations until you find one that is identical to the simplified version.

The answer is the second one: 8x²y + 7x² - 5y² + 3x - 2 + 2x²y - x² - 2y² - 3x - 4

6 0

3 years ago

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In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen

Mashcka [7]

Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

4 0

3 years ago

Help really fast round 93.0129 to the nearest thousandth

maksim [4K]

The thousandth place is the 3rd digit after the decimal.

So your answer is 93.013

3 0

3 years ago

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