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Ksenya-84 [330]

2 years ago

6

Help me pls with my geometry

1 answer:

garik1379 [7]2 years ago

5 0

Answer:

answer in pic above

Step-by-step explanation:

did this over thrice before coming to the right answer lol

i hope it helps you xxx

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What's 30.08 divided by 4

belka [17]

30.08 divided by 4 is 7.52

6 0

3 years ago

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The valume of (3-5)4(2)-16+2

rusak2 [61]

Answer:

-30

Step-by-step explanation:

Calculate the difference:

(3-5)x4x2- 16 plus 2

Result: -2x4x2- 16 plus 2

calculate the product...

Result: -16-16 plus 2

Calculate the sum or difference

ANSWER: -30

8 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

6 0

2 years ago

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Given h(t) = -2(t+ 5)^2 +4, find h(-8).

bekas [8.4K]

Answer:

Substitute -8 as x into the equation.

h(-8)=-2(-8+5)^2+4

h(-8)=-2(-3)^2+4

h(-8)=-2(9)+4

h(-8)=-18+4

h(-8)=-14

:)

7 0

3 years ago

How to add fractions with different denominators and negative numbers?

Kisachek [45]

Let find the least of common multiple = LCM it’s for the denominators.
Multiple of the numerator then the denominator to get the denominators
Don’t forget to to add the numerator but leave the denominators the same

8 0

3 years ago