MATHEMATICS COLLEGE

Rewrite 3/11 amd 1/4 so they have a common denominator

Answers

Answer 1

Answer:

Hello!

Answer:

$\Large \boxed{\sf (12)/(44) ~~ and~~ (11)/(44) }$

Step-by-step explanation:

We want that the fractions 3/11 and 1/4 have a common denominator.

Let's find the LCM (least common multiple) of 4 and 11:

$\text{\sf Multiples of 4:}~~ \sf 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, \boxed{\sf 44}, 48\n\n\text{\sf Multiples of 11:} ~~ \sf 11, 22, 33, \boxed{\sf 44}, 55, 66, 77, 88, 99, 110$

So the LCM of 4 and 11 is 44.

Convert fractions over 44:

$\sf (3)/(11) = (3 * 4 )/(11 * 4) = \boxed{\sf (12)/(44)}$

$\sf (1)/(4) = (1 * 11 )/(4 * 11) = \boxed{\sf (11)/(44)}$

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How many pounds are there in a 72-ounce package of ground beef?(1 pound=16 ounces)

Answers

Answer:

4.5 pounds of ground beef

Answer:

4.5

Step-by-step explanation:

You would divide 72 by 16 to get 4.5 as the answer.

Evaluate the expression b/2-3 if b=8.

Answers

Answer:

-8

Step-by-step explanation:

b/2-3

substitute 8 for b

8/2-3

then solve

8/2-3

8/-1

-8

Suppose we express the amount of land under cultivation as the product of four factors:Land = (land/food) x (food/kcal) x (kcal/person) x (population)

The annual growth rates for each factor are:
1. the land required to grow a unit of food, -1% (due to greater productivity per unit of land)
2. the amount of food grown per calorie of food eaten by a human, +0.5%
3. per capita calorie consumption, +0.1%
4. the size of the population, +1.5%.

Required:
At these rates, how long would it take to double the amount of cultivated land needed? At that time, how much less land would be required to grow a unit of food?

Answers

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following annual growth rates:

land/food = - 1%

food/kcal = 0.5%

kcal/person = 0.1%

population = 1.5%

Σ annual growth rates = (-1 + 0.5 + 0.1 + 1.5)% = 1.1% = 0.011

Exponential growth in Land :

L = Lo * e^(rt)

Where Lo = Initial ; L = increase after t years ; r = growth rate

Time for amount of cultivated land to double

L = 2 * initial

L = 2Lo

2Lo = Lo * e^(rt)

2 = e^(0.011t)

Take the In of both sides

In(2) = 0.011t

0.6931471 = 0.011t

t = 0.6931471 / 0.011

t = 63.01 years

Land per unit of food at t = 63.01 years

L = Fo * e^(rt)

r = growth rate of land required to grow a unit of food = 1% = 0.01

L/Fo = e^(-0.01* 63.01)

L/Fo = e^(−0.6301)

= 0.5325385 = 0.53253 * 100% = 53.25%

Land per unit now takes (100% - 53.25%) = 46.75%

Need answer please !!!!!

Answers

Answer: -9

Step-by-step explanation:

f(x)=-5x-4

f(1)=-5(1)-4

=-5-4

f(1) =-9

Use the exponential growth model, A = A0 e^kt to show that the time is takes a population to double (to frow from A0 to 2 A0) is given by t = ln 2/k.

Answers

Answer:

Proof below

Step-by-step explanation:

Exponential Grow Model

The equation to model some time dependant event as an exponential is

$A=A_oe^(kt)$

Where Ao is the initial value, k is a constant and t is the time. With the value of Ao and k, we can compute the value of A for any time

We are required to find the time when the population being modeled doubles from Ao to 2 Ao. We need to solve the equation

$2A_o=A_oe^(kt)$

Simplifying by Ao

$2=e^(kt)$

Taking logarithms in both sides

$ln2=lne^(kt)$

By properties of logarithms and since lne=1

$ln2=kt\cdot lne=kt$

Solving for t

$\displaystyle t=(ln2)/(k)$

Hence proven

The speed of light travels at a rate of approximately 3.0 * 10^8 meters per second. How far, in meters, does light travel in 10^5 seconds?

Answers

Answer:

Step-by-step explanation:

There is a few ways to go about doing this problem. I am going to setup a function.

x = seconds

f(x) = distance = How far

f(x) = (3.0* 10^8) x

Insert 10^5 where x is located

(3.0* 10^8) x

Multiply

(3.0* 10^8) (10^5)

Do the math

(3.0* 10^8) (10^5) = 3 * 10^13

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Is 4/12 closer to 0 , 1 or 1/2