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Ksenya-84 [330]
2 years ago
6

Help me pls with my geometry

Mathematics
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
Read 2 more answers
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
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