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4 years ago

BRAINLIEST!! LOTS of POINTS!!!! HURRY HURRY

Mathematics

2 answers:

Semenov [28]4 years ago

8 0

Answer/Step-by-step explanation:

<u>Steps to solve: 2/9 − (−1/3 ÷ 3/5)</u>

(-1/3 ÷ 3/5)

= (-1/3 x 5/3)

= (-5/9)

-2/9 - (-5/9)

= -2/9 + 5/9

= 3/9

<u>To check:</u>

5/9 - 2/9 = 3/9

-Dominant- [34]4 years ago

6 0

Answer:

Step-by-step explanation:

−2/9−(−1/3÷3/5)

=-2/9-[(-1/3)*(5/3)]

=−2/9−[−1*5/3*3]

=−2/9−(−5/9)

=(-2+5)/9

=3/9

=1/3

As the question says "Enter your answer, as a fraction in simplest form", 3/9 is NOT the right answer.

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45 ft²

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Write an equation for the line?

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Answer:

y = 2x + 3

Step-by-step explanation:

Given:

(2, 7) and (0, 3)

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3 years ago

Jackson used all the butter to make the brownies. If each brownie is 1/9 of a batch, how many brownies did he make? Enter your a

Basile [38]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

F(x) = -x times 9 +1 for x=-3

sineoko [7]

Substitute the given value into the function and evaluate.
The answer is -10

6 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

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3 years ago

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