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bearhunter [10]
3 years ago
11

I

Mathematics
1 answer:
Ivanshal [37]3 years ago
5 0

Answer:

4,000 mm

Step-by-step explanation:

4 x 1,000= 4,000

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Help!!!How does the slope of g(x) compare to the slope of f(x)?
attashe74 [19]

we know that

The formula to calculate the slope m between two points is equal to

m=\frac{(y2-y1)}{(x2-x1)}

Step 1

<u>Find the slope of the function f(x)</u>

Let

A( 0,-2)\\B(1,0)

substitute the values in the formula  above

m=\frac{(0+2)}{(1-0)}

m=2

<u>the slope of the function f(x) is m=2</u>

Step 2

<u>Find the slope of the function g(x)</u>

Let

C( -4,0)\\D(0,2)

substitute the values in the formula  above

m=\frac{(2-0)}{(0+4)}

m=\frac{2}{4}

m=\frac{1}{2}

the slope of the function g(x) is m=\frac{1}{2}

therefore

slope f(x) > slope g(x)

or

slope g(x) < slope f(x)

<u>the answer is the option </u>

B.The slope of g(x) is less than the slope of f(x).

4 0
3 years ago
Read 2 more answers
Answer these questions with work
sweet-ann [11.9K]
First, you make 9 into a fraction which would become 36/4
then, subtract 9/4 from both sides so that would give you 3/4x = 27/4
then divide each side by 3/4 (same thing as multiply by its reciprocal) and that would give you x=9
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What is the answer to 5 and 3/4 - 2 1/4
sergey [27]

Answer:

<h2>The answer you are looking for is 3 and 1/2</h2><h2>(or 7/2 or 3.5)</h2>

Step-by-step explanation:

<h2>plz mark as brainliest!!!</h2>
8 0
3 years ago
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1. Consider the population model dP dt = 0.2P 1 − P 135 , where P(t) is the population at time t. (a) For what values of P is th
Kryger [21]

Answer:

a

  P = 0  OR  P = 135

b

P > 0 and P < 135

OR

P > 0 and P < 135

c

Generally the carrying capacity is can be defined as the highest amount of population and environment can support for an unlimited duration or time period

d

P = 67.5

Step-by-step explanation:

From the question we are told that

The population model is \frac{dP}{dt}  =  0.2P(1 - \frac{P}{135} )

Generally at equilibrium

\frac{dP}{dt} = 0

So

0.2P = 0

=> P = 0

Or

(1 - \frac{P}{135} ) = 0

=> P = 135

Thus at equilibrium P = 0 or P = 135

Generally when the population is increasing we have that

\frac{dP}{dt} > 0

So

0.2P > 0

=> P > 0

and

(1 - \frac{P}{135} ) > 0

P < 135

Now when the first value of P i.e P< 0 for \frac{dP}{dt} > 0

P_2 > 135

So when population increasing the values of P are

P > 0 and P < 135

OR

P > 0 and P < 135

So to obtain initial values of P where the population converge to the carrying capacity as t \to [\infty]

The rate equation can be represented as

\frac{dP}{dt}  =  \frac{1}{5}P (1 - \frac{P}{135} )

So we will differentiate the equation again we have that

\frac{d^2 P}{dt^2} = \frac{(1 - \frac{P}{135} )}{5}  - \frac{P}{675}

Now as  t \to [\infty]

\frac{d^2 P}{dt^2} \to  0

So

   \frac{(1 - \frac{P}{135} )}{5}  - \frac{P}{675} = 0      

=>    \frac{(1 - \frac{P}{135} )}{5}   =  \frac{P}{675}

=> P = 67.5

5 0
3 years ago
A new drug on the market is known to cure 20% of patients with breast cancer. If a group of 20 patients is randomly
Alenkasestr [34]

Answer:

It's actually C

Step-by-step explanation:

don't forget about the probability of 0 too

you have to add the two probability formulas

4 0
3 years ago
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