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2 years ago

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1 answer:

Triss [41]2 years ago

8 0

Answer:

$T_1 = 8$

$T_2 = 14$

$T_3 = 22$

Step-by-step explanation:

Given

$T_n = n^2 + 3n + 4$

Required:

Find the first 3 terms

To do this, we take the values of n to be 1,2 and 3.

For the first term; n = 1

$T_n = n^2 + 3n + 4$

$T_1 = 1^2 + 3*1 + 4$

$T_1 = 1+ 3+ 4$

$T_1 = 8$

For the second term; n = 2

$T_n = n^2 + 3n + 4$

$T_2 = 2^2 + 3 * 2 + 4$

$T_2 = 4 + 6+ 4$

$T_2 = 14$

For the third term; n = 3

$T_n = n^2 + 3n + 4$

$T_3 = 3^2 + 3 * 3 + 4$

$T_3 = 9 + 9 + 4$

$T_3 = 22$

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A vendor sold a combined 234 sodas and hot dogs. The number of sodas was two times the number of hot dogs sold. Find the number

babymother [125]

Answer:

156 hot dogs
78 sodas

Step-by-step explanation:

A group of 2 hot dogs and 1 soda makes a group of 3 items. There were ...

234/3 = 78

groups of items sold.

Since there is 1 soda in the group, that's 78 sodas. The number of hot dogs is 2 times that, or 156 hot dogs.

78 sodas and 156 hot dogs were sold.

6 0

2 years ago

Find the inequality represented by the graph.

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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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Identify each of the following as rational or irrational

Lera25 [3.4K]

√23-irrational
104.42-rational
√64-irrational
49.396-rational
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1 year ago

I need help with this one

OlgaM077 [116]

Answer:

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Step-by-step explanation:

Let's work backwards:

Circumference = 2πr

(37.68)/2π = r

Area = π $r^{2}$

Area = π * $((37.68)/2pi)^{2}$

Area = 112.983 square units

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3 years ago

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