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aleksklad [387]

3 years ago

14

Please help me, Please I will give you 15 points and brainliest

2 answers:

maksim [4K]3 years ago

8 0

Answer:

Sample Response: The inequality could be t > 900. Draw a number line so it includes 900. Draw an open circle on 900. Shade all the numbers greater than 900, or to the right.

Step-by-step explanation:

stealth61 [152]3 years ago

4 0

Answer: The inequality could be t > 900. Draw a number line so it includes 900. Draw an open circle on 900. Shade all the numbers greater than 900, or to the right.

Step-by-step explanation:

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Simplify the expression 6[8+8](3+7)] ?

krok68 [10]

Answer:

960

Step-by-step explanation:

6[8+8](3+7)

Parentheses first

6[16] (10)

Then multiply from left to right

96*10

960

7 0

3 years ago

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A recipe for fruit salad uses 3 times as many apples as oranges. drag apples to represent the number of apples used for every or

Serga [27]

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3 0

2 years ago

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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

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

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Mr.Gleason's math class lasts 40 minutes.Math class starts at 9:55 A.M. At what time does math class end?

klio [65]

Math class ends at 10:35 am

5 0

3 years ago

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Another question! Help if you can

ioda

The answer is D because it can only have one symmetrical line straight down the middle.

3 0

3 years ago