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1 year ago

13

I will give branniest

1 answer:

KonstantinChe [14]1 year ago

8 0

Answer:

112.5

Step-by-step explanation:

24x=2700

=> x= 2700/24

= 112.5

Therefore, x= 112.5

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Which trigonometric function would you use to solve the problem?

umka21 [38]

you would use tan or cot

3 0

3 years ago

0.750 0.835 0.955 1.205 witch of this numers have a 5 in the hundreths place

MrRissso [65]

0.750,and 0.955 because both have 5s in the hundredths place

4 0

3 years ago

Find the values of f(x) = 8x²-1 for the domain values {-2,5} (input answers from least to greatest}

Whitepunk [10]

Answer:

So i just used the calculator to help me find the two co exisiting lines and then worked my way up to find the nearest decimal to fit the equation for the missing parablar length which was parallel to find the cointerior angle which lead me to get a decimal number involving an ordinary form number.

7 0

2 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

What is the slope of (-3,5) and (2,0)

IrinaVladis [17]

Answer:

is there anymore to the question?

Step-by-step explanation:

8 0

2 years ago