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2 years ago

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Leo drew a line that is perpendicular to the line shown on the grid and passes through point (F. G). Which of the following is t

he equation of Leo's line?
Oy-G=-2(x-F)
Oy+F=2(x+G)
Oy+F=1/2(x+G)
Oy-G=1/2( x-F)

1 answer:

tensa zangetsu [6.8K]2 years ago

7 0

The equation of the perpendicular line drawn by Leo is $y-G=\dfrac{-1}{2}(x-F)$ . Option C is the correct answer.

<h3>How to determine the equation of a line?</h3>

A line is drawn perpendicular to the line shown in the image. The perpendicular line passes through the coordinate point (F,G).

The slope of the line from the graph is-

$m=\dfrac{y-intercept}{x-intercept}\\\\\\m=2$

Therefore, the slope of the perpendicular line is $\dfrac{-1}{2}$ .

Also, it is being given that Leo's line is passing through the coordinate point .

So, the equation of the Leo's line is-

$y-G=\dfrac{-1}{2}(x-F)$

Thus, the equation of the perpendicular line drawn by Leo is .

$y-G=\dfrac{-1}{2}(x-F)$

Learn more about the equation of line here- brainly.com/question/20632687

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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

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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

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

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

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

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

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

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

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