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faltersainse [42]

3 years ago

5

If angles are supplementary, then one of the angles is an obtuse angle.

1 answer:

oksian1 [2.3K]3 years ago

8 0

Always, because supplementary means it would add up to 180, but one of the angles must be greater than 90 otherwise if its less and the other one is less, then they will not add to 180

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Please solve<br>I really need help

Llana [10]

9514 1404 393

Answer:

x = 12

Step-by-step explanation:

The similarity statement tells you corresponding sides have the same ratio.

AB/BC = FE/EC

x/10 = 6/5

x = 10(6/5)

x = 12

7 0

3 years ago

Find the value of 2m for m= 1,2 and 3

Nonamiya [84]

Use the substitution method

Answers:

2m

2(1)

= 2

2(2)

= 4

2(3)

= 6

7 0

3 years ago

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I don't know how to do these sort of problems... Can someone help me out?

VLD [36.1K]

180=76+4x
180-76=4x
104=4x
104/4=x
26=x
Measurement A= 26

6 0

3 years ago

Someone please answer !!!!

Goryan [66]

The answer is C

By the way, that expression means "n choose k". Hope I helped!

7 0

3 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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