What is not an undefined term

Solve the following system of linear equations {2x-7y=10 {5x -6y=2

Sliva [168]

2x-7y=10 = $\frac{2}{7}$

5x -6y=2 = $\frac{5}{6}$

5 0

3 years ago

a fence is to be built to enclose a rectangular area of 240 sq ft. the fence along three sides is to be made of material that co

bixtya [17]

Answer:14400

Step-by-step explanation:

8 0

3 years ago

Country A has a growth rate of 4.9% per year. The population is currently 4 comma 151,000, and the land area of Country A is 1

kvv77 [185]

Answer:

There will be one person on 1 square yard of land after 1,892,147.588 years.

Step-by-step explanation:

Continuous exponential growth formula:

$P(t)=Pe^{rt}$

P(t)= Population after t years.

P= Initial population

r=rate of growth.

t= time in year

Given that,

Growth rate of country A (r)= 4.9% per year=0.049 per year.

Initial population (P)= 151,000.

Land area of country area= 14,000,000,000 square yards.

There will be one person on one square yard of land.

So, there will be 14,000,000,000 person for 14,000,000,000 square yard of land in country A.

P(t)=14,000,000,000 person

$\therefore 14,000,000,000= 151,000 e^{0.049t}$

$\Rightarrow e^{0.049t}=\frac{ 14,000,000,000}{ 151,000}$

Taking ln both sides

$\Rightarrow ln|e^{0.049t}|=ln|\frac{ 14,000,000,000}{ 151,000}|$

$\Rightarrow {0.049t}=ln|\frac{ 14,000,000,000}{ 151,000}|$

$\Rightarrow t}=\frac{ln|\frac{ 14,000,000,000}{ 151,000}|}{0.049}$

$\Rightarrow t}=1,892,147.588$ years

There will be one person for every square yard of land after 1,892,147.588 years.

7 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

what interest rate is the local credit union giving you on a car loan of 25,000 for 5 years if they will collect 1500 in interes

lianna [129]

Answer:

1.2%

Step-by-step explanation:

4 0

2 years ago