All categories

3 years ago

Identify the equation in slope-intercept for for the line containing the point (-3,5) and patella to y = -2/3x+5/3

Mathematics

1 answer:

Olegator [25]3 years ago

8 0

Answer:

Y = - 2/3 x + 3

Step-by-step explanation:

y = -2/3x+5/3

Slope = -2/3. .. parallel line has same slope

For (- 3 , 5)

Y intercept = y - my = 5 - ( (-2/3) x (-3) ) = 3

Y = slope x X + y intercept

Y = - 2/3 x + 3

You might be interested in

52÷2= ?? find the quotient and remainder

vodomira [7]

52/2= 26

The quotient is 26, but there are no remainder.

Hope this helps!

7 0

3 years ago

Read 2 more answers

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

6 0

2 years ago

Read 2 more answers

Please help.

Ad libitum [116K]

Answer:

In what article are you most likely to find the sentence “The coma streamed out until it formed a tail of dust and light”?

“Comets in the Sky”

“How to Punctuate Sentences”

“Water Under the Bridge”

“Medical Histories”

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

A junior college is going to choose a Homecoming Queen, and the candidates are the Freshman Queen and the Sophomore Queen. If th

KonstantinChe [14]

Count the votes, counting each sophomore ballot as 1.5 votes and each freshmen ballot as 1 vote.
ur doing this because there is 200 more freshmen then sophomores...and if u count each sophomore vote as 1.5, it would make up for the 200 more freshmen

5 0

2 years ago

Is 86, 75, 97, 58. 94, and 58, odd or even

alexira [117]

This is even because they all can be divided if it was odd you would not be able to decide them

5 0

3 years ago