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3 years ago

What is the equation of the line that has a slope of -4 and goes through the point (4,-6)

Mathematics

1 answer:

Elis [28]3 years ago

6 0

It would be Y=-4x+10. Please mark me brainliest. It’s fine if you don’t want to tho.

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Me ayudan por favor con estos ejercicios de álgebra con la 15, 16 y 17

SashulF [63]

No soy muy buena con álgebra pero te recomiendo esta app que se llama “Symbolab” donde solo tienes que escanear la tarea y te da la respuesta.

7 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

Hi I really need help with this question please,

Mrrafil [7]

Answer:

${ ( \sqrt{30 - x} )}^{2} = {x}^{2} \\ 30 - x = {x}^{2} \\ \\ {x}^{2} + x - 30 = 0 \\ delta = 1 - 4 \times ( - 30) \times 1 = 121 \\ x1 = \frac{ - 1 + \sqrt{121} }{2 \times 1} = \frac{ - 1 + 11}{2} = 5 \\ x2 = \frac{ - 1 - \sqrt{121} }{2 \times 1} = \frac{ - 1 - 11}{2} = - 6 \\$