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3 years ago

Connie started with___ in her saving account Fill in the blank

Mathematics

2 answers:

Taya2010 [7]3 years ago

5 0

Answer:

Step-by-step explanation:

Irina18 [472]3 years ago

3 0

Answer:

Connie started with <u>$25</u> in her savings account.

Step-by-step explanation:

The y-intercept is the value of the function at time zero.

At x = 0, which is week zero, when Connie opened the account, the y value is 25. That means she opened the account with $25.

Answer: $25

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Find the volume of a right circular cone that has a height of 17.6 m and a base with a

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Answer:

1179m^3

Step-by-step explanation:

Given radius=8m

Height=17.6m

Volume of the cone=1/3πr^2h

1/3 πr^2h

=1/3×3.14×8×8×17.6

=1178.96

Nearest tenth is 1179

So volume is 1179m^3

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2 years ago

You bought items at Best Buy that rang up $143.75. You have a 30% discount and have to pay 9.5% sales tax.

Mandarinka [93]

Answer:

$92.33

Step-by-step explanation:

You take 143.75 and multiply by .3 to find the amount you will discount from 145.75, to get 102.025. Then multiply that by .095 to get 9.692375. Subtract that from 102.025 to get 92.33.

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3 years ago

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Product of square of x and cube of y

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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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2 years ago

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