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Svetlanka [38]
3 years ago
12

True or False:

Mathematics
1 answer:
Naya [18.7K]3 years ago
6 0

we can check if a given sequence is geometric or not by checking the terms present in it ,

Let us find ratio of each term with it's previous term

if that ratio comes out same for every two consecutive terms then it is a geometric sequence , otherwise it is not geometric sequence .

Let us work on first one : given is : -1,1,-1,1.............

second term/ first term = 1/-1 = -1

third term/ second term = -1/1 = -1

fourth term / third term = 1/-1 = -1

fifth term/ fourth term = -1/1 = -1

because that ratio is same so it is a geometric sequence .

* let us now work with second sequence .

eE

second term/ first term = -6/-2 = 3

third term/ second term = -18/ -6 = 3

fourth term / third term = -54 / -18 = 3

hence this is also geometric sequence .

Answer : both are geometric sequence .

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The correct answer is 105.8
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Find the missing value so that the line passing through the points has the given slope.
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Answer:

11

Step-by-step explanation:

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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

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Hello!
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