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2 years ago

2.) Use the Slope Intercept Form of a line to find the equation of the line from point C to point D.

Mathematics

2 answers:

mezya [45]2 years ago

8 0

First we need slope

C=(0,0)
D(7,12)

$\\ \sf\longmapsto m=\dfrac{12-0}{7-0}$

$\\ \sf\longmapsto m=\dfrac{12}{7}$

Put D co-ordinates on y=mx+b

$\\ \sf\longmapsto 12=\dfrac{12}{7}(7)+b$

$\\ \sf\longmapsto 12=12+b$

$\\ \sf\longmapsto b=12-12$

$\\ \sf\longmapsto b=0$

Now

slope intercept form.

$\\ \sf\longmapsto y=\dfrac{12}{7}x$

As b=0

mrs_skeptik [129]2 years ago

3 0

Answer:

• General equation of a line:

${ \rm{y = mx + b}}$

Consider end points of line CD: (0, 0) and (7, 12):

• find the slope, m:

${ \rm{slope = \frac{y _{2} - y _{1} }{x _{2} - x _{1} } }} \\ \\ { \tt{m = \frac{12 - 0}{7 - 0} }} \\ \\ { \underline{ \tt{ \: \: m = \frac{12}{7} \: \: }}}$

• using end point (0, 0), find the y-intercept, b:

${ \rm{y = mx + b}} \\ \\ { \rm{0 = ( \frac{12}{7} \times 0) + b}} \\ \\ { \tt{b = 0}}$

• therefore, equation of line CD is;

${ \boxed{ \boxed{ \tt{ \: \: y = \frac{12}{7}x \: \: }}}}$

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Read 2 more answers

By 3:00 pm, the temperature increased by another 5 degrees Fahrenheit was the temperature at 3:00pm positive or negative?

Ad libitum [116K]

Answer:

Part a) The temperature at noon was -3°F

Part b) The temperature at 3:00 pm was +2°F

Part c) The temperature at 11:00 pm was -6°F

Step-by-step explanation:

The complete question is

When Leo woke up he saw that the temperature was -8°F. By noon the temperature has increased 5°F. Part a) What was the temperature at noon?

Part b) By 3:00 pm, the temperature increased by another 5 degrees Fahrenheit was the temperature at 3:00 pm positive or negative?

Part c) By 11:00 pm, the temperature had dropped 8°F. Was the

temperature at 11:00 PM positive or negative? Explain.

Part a) we know that

You can use a number line to adds the numbers

-8+5=-3°F

therefore

The temperature at noon was -3°F

Part b) we know that

the temperature increased by another 5 degrees

You can use a number line to adds the numbers

-3+5=+2°F

therefore

The temperature at 3:00 pm was +2°F

Part c) we know that

the temperature had dropped 8°F

You can use a number line to adds the numbers

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2 years ago

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3 years ago

A. Do some research and find a city that has experienced population growth.

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A. The city we will use is Orlando, Florida, and we are going to examine its population growth from 2000 to 2010. According to the census the population of Orlando was 192,157 in 2000 and 238,300 in 2010. To examine this population growth period, we will use the standard population growth equation $N_{t} =N _{0}e^{rt}$
where:
$N(t)$ is the population after $t$ years
$N_{0}$ is the initial population
$t$ is the time in years
$r$ is the growth rate in decimal form
$e$ is the Euler's constant
We now for our investigation that $N(t)=238300$ , $N_{0} =192157$ , and $t=10$ ; lets replace those values in our equation to find $r$ :
$238300=192157e^{10r}$
$e^{10r} = \frac{238300}{192157}$
$ln(e^{10r} )=ln( \frac{238300}{192157} )$
$r= \frac{ln( \frac{238300}{192157}) }{10}$
$r=0.022$
Now lets multiply $r$ by 100% to obtain our growth rate as a percentage:
$(0.022)(100)$ =2.2%
We just show that Orlando's population has been growing at a rate of 2.2% from 2000 to 2010. Its population increased from 192,157 to 238,300 in ten years.

B. Here we will examine the population decline of Detroit, Michigan over a period of ten years: 2000 to 2010.
Population in 2000: 951,307
Population in 2010: 713,777
We know from our investigation that $N(t)=713777$ , $N_{0} =951307$ , and $t=10$ . Just like before, lets replace those values into our equation to find $r$ :
$713777=951307e^{10r}$
$e^{10r} = \frac{713777}{951307}$
$ln(e^{10r} )=ln( \frac{713777}{951307} )$
$r= \frac{ln( \frac{713777}{951307}) }{10}$
$r=-0.029$
$(-0.029)(100)$ = -2.9%.
We just show that Detroit's population has been declining at a rate of 2.2% from 2000 to 2010. Its population increased from 192,157 to 238,300 in ten years.

C. Final equation from point A: $N(t)=192157e^{0.022t}$ .
Final equation from point B: $N(t)=951307e^{-0.029t}$
Similarities: Both have an initial population and use the same Euler's constant.
Differences: In the equation from point A the exponent is positive, which means that the function is growing; whereas, in equation from point B the exponent is negative, which means that the functions is decaying.

D. To find the year in which the population of Orlando will exceed the population of Detroit, we are going equate both equations $N(t)=192157e^{0.022t}$ and $N(t)=951307e^{-0.029t}$ and solve for t:
$192157e^{0.022t} =951307e^{-0.029t}$
$\frac{192157e^{0.022t} }{951307e^{-0.029t} } =1$
$e^{0.051t} = \frac{951307}{192157}$
$ln(e^{0.051t})=ln( \frac{951307}{192157})$
$t= \frac{ln( \frac{951307}{192157}) }{0.051}$
$t=31.36$
We can conclude that if Orlando's population keeps growing at the same rate and Detroit's keeps declining at the same rate, after 31.36 years in May of 2031 Orlando's population will surpass Detroit's population.

E. Since we know that the population of Detroit as 2000 is 951307, twice that population will be $2(951307)=1902614$ . Now we can rewrite our equation as: $N(t)=1902614e^{-0.029t}$ . The last thing we need to do is equate our Orlando's population growth equation with this new one and solve for t:
$192157e^{0.022t} =1902614e^{-0.029t}$
$\frac{192157e^{0.022t} }{1902614e^{-0.029t} } =1$
$e^{0.051t} = \frac{1902614}{192157}$
$ln(e^{0.051t} )=ln( \frac{1902614}{192157} )$
$t= \frac{ln( \frac{1902614}{192157}) }{0.051}$
$t=44.95$
We can conclude that after 45 years in 2045 the population of Orlando will exceed twice the population of Detroit.

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3 years ago

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bulgar [2K]

Answer:

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Step-by-step explanation:

We require 2 equations with the repeating digits (63) placed after the decimal point.

let x = 0.636363..... (1) multiply both sides by 100

100x = 63.6363... (2)

Subtract (1) from (2) thus eliminating the repeating digits

99x = 63 ( divide both sides by 99 )

x = $\frac{63}{99}$ = $\frac{7}{11}$ ← in simplest form

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2 years ago