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2 years ago

15

If a line goes through the following two points: (0,-2) and (3,4), the slope of the line is:

1 answer:

atroni [7]2 years ago

6 0

Answer:

y = 2x – 2

Step-by-step explanation:

You might be interested in

How to convert 45% into a fraction

JulsSmile [24]

Answer:

$\frac{9}{20}$

Step-by-step explanation:

Per cent means ' out of a hundred' , thus

45% = $\frac{45}{100}$ ← divide ( cancel) both 45 and 100 by 5

45% = $\frac{9}{20}$ ← in simplest form

8 0

3 years ago

F(x)=2x 2 +5√ (x+2) f(2)=

irga5000 [103]

f(x) 2x² + 5√(x+2)

f(2) = 2(2)² + 5√(2+2)

= 2(4) + 5√4

= 8 + 10

= 18 <======= answer

4 0

3 years ago

Answer for fee rbux and branlest!!!! i need answer NOW or i will be DIE (not good!!!)

pshichka [43]

Answer:

its B homie

Step-by-step explanation:

U stuped

8 0

3 years ago

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

36÷(6÷3) plz helppppooooop

Sunny_sXe [5.5K]

Answer:

18

Step-by-step explanation:

Divide with by paratheses first. Use PEMDAS method

7 0

3 years ago

Read 2 more answers