MATHEMATICS HIGH SCHOOL

a population of 1200 leopards decreases by 12% per year. How many leopards will there be in the population after 6 years? Round your answer to the nearest whole number.

Answers

Answer 1

Answer:

Answer:

557

Step-by-step explanation:

We are given that a population of 1200 leopards decreases by 12% per year.

So, we will use exponential function: $a_n=a_0(1-r)^x$

Where $a_n$ is the population left.

$a_0$ is the initial population.

r = rate of decrease

x = time

Now $a_0=1200$

r =12%=0.12

x = 6 years

Substitute the values.

$a_n=1200(1-0.12)^6$

$a_n=1200(0.464404086784)$

$a_n=557.28$

Thus there will be 557 leopards in the population after 6 years.

Answer 2

Answer: After 6 years approximately 557 leopards would be left.

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A.Tell if the sequence is arithmetic, geometric orneither: 5, 30, 180, 1080
Carrots sell for $2.10 per pound, and crackers sell for $2.90 per pound. Glen bought some carrots and some crackers. The total weight was 2.3 pounds and cost $6.03.How many pounds of carrots and how many pounds of crackers did Glen buy? A.0.5 pounds of carrots; 1.8 pounds of crackers B.0.8 pounds of carrots; 1.5 pounds of crackers C.1 pound of carrots; 1.3 pounds of crackers D.1.5 pounds of carrots; 0.8 pound of crackers

Identify the transformation.

Answers

Answer:

C) translation 5 united left, 1 unit up

Step-by-step explanation:

to figure this out you can take one point and translate it into 5 units left and 1 up and it will be in the spot of the new figure, Reflection and rotation are not correct because the figure would be oriented differently, and it is not D because the figure with ' is the new transformation

16 men are employed to do a work in 20 days. at the end of 12 days the work is only half done. how many additional number of men should be employed to complete the work in the stipulated time?

Answers

16 men in 12 days can do 0.5work
1 man 1 day can do 0.5/(16x12) of a work

8days to do 0.5 of a work
1 man 8 days can do 8x0.5/(16x12) = 0.5x(1/24) of a work
... if you notice we need 24 men

... more detail
in8days, 0.5x(1/24) of a work need 1 man
in8days, 0.5 of a work need 0.5/(0.5x(1/24)) = 24 men

point is to find 'amount of work done per man per unit of time' then work out from that.

Can 92 be classified as a rational number

Answers

Yes, because 92 is an integer and a whole number.

Yes. A Rational Number is a real number that can be written as a simple fraction (i.e. as a ratio). Whole numbers, terminating decimals, and fractions are considered rational.

15x—3y=12
Y=5x—4

4x—y=-4
-8x+2y=2

Answers

15x-3y=12
y=5x—4
I'm also not sure what your question is, but this is what I got by solving it like a regular equation.

15x-3(5x-4)=12
15x-15x+12=12
0=12-12
0=0

4x-y=-4
-8x+2y=2

-y=-4-4x /: (-1)
y=4+4x

-8x+2y=2
-8x+2(4+4x)=2
-8x+8+8x=2
0=-6

A house cost $120,000 when it was purchased. The value of the house increases by 10% each year. Find the rate of growth each month.

Answers

FIRST MODEL:

Well the model for the value of the house is:

$V={ \left( \frac { 11 }{ 10 } \right) }^( t )\cdot 120000$

V = Value

t = Years passed {t≥0}

-----------------------

When t=0, V=120000

When t=1, V=132000

When t=2, V=145200

etc... etc...

---------------------------

Now, this model is actually curved so there is no constant rate of growth each month. We can only calculate what the rate of growth is at a particular time. If we want to find out the rate of growth at a particular time, we must differentiate the formula (model) above.

--------------------------

$V={ \left( \frac { 11 }{ 10 } \right) }^( t )\cdot 120000\n \n \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^( t )\cdot 120000 \right) } }$

$\n \n \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^( t ) \right) } } +\ln { \left( 120000 \right) } \n \n \ln { V=t\ln { \left( \frac { 11 }{ 10 } \right) } } +\ln { \left( 120000 \right) }$

$\n \n \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \n \n V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\n \n \therefore \quad \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^( t )\cdot 120000$

Plug any value of (t) that is greater than 0 into the formula above to find out how quickly the investment is growing. If you want to find out how quickly the investment was growing after 1 month had passed, transform t into 1/12.

The rate of growth is being measured in years, not months. So when t=1/12, the rate of growth turns out to be 11528.42 per annum.

SECOND MODEL (What you are ultimately looking for):

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

V = Value of house

t = months that have gone by {t≥0}

Formula above differentiated:

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000\n \n \ln { V } =\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000 \right) }$

$\n \n \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } } \right) } } +\ln { \left( 120000 \right) }$

$\n \n \ln { V=\frac { t }{ 12 } } \ln { \left( \frac { 11 }{ 10 } \right) } +\ln { \left( 120000 \right) }$

$\n \n \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) }$

$\n \n V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\n \n \therefore \quad \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

When t=1, dV/dt = 960.70 (2dp)

dV/dt in this case will measure the rate of growth monthly. As more money is accumulated, this rate of growth will rise. The rate of growth is constantly increasing as the graph of V is actually a curve. You can only find out the rate at which the house value is growing monthly at a particular time.

Which equation represents the line whose slope is 2 and whose y-intercept is 6

Answers

The basic format for questions like these are y=mx+b where m stands for the slope and the b stands for the y-intercept, and x and y stand for the coordinates of the equation. So if you fill it in, it would be y=2x+6.

slope of 2 means that the the x part is 2x

So when x is 0 y must equal 6

so the equation is y = 2x + 6

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