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2 years ago

6

Which functions are nonlinear? Select TWO that apply.

2 answers:

den301095 [7]2 years ago

4 0

E and A I think or it might be E and D

Valentin [98]2 years ago

3 0

Answer:

D and E

Step-by-step explanation:

Think about what nonlinear means. "Non" means not, and "Linear" means a line, so you can pick choice E

D is also non-linear because there isn't a constant change in both the y-variable and x-variable.

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Pleas help asap. Thank you<br><br> A.53°<br><br> B.39°<br><br> C.37°<br><br> D.27°

Viktor [21]

You will use the functions soh cah toa. you know that angle b is 90 so the other 2 angles have to add up to 90°. with the Pythagorean theorem we can get that cb is 12 feet. so we now have all side lengths and 1 angle. you can pick out that 53° will be your final answer after doing the basic functions!

4 0

4 years ago

The octagon is a regular polygon. What is the measure of each interior angle in the polygolon

MrRissso [65]

(N-2)*180/n
Octagon has 8 sides
(8-2)*180/8
=135

5 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

3 years ago

Read 2 more answers

Alan, Bobby and Chris are comparing their ages. The ratio of Alan's age to

SOVA2 [1]

Answer:

1638392772127273930337733930282823829393

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3 years ago

Renuka has two pounds of pepper in her cupboard. She knows that there are 16 ounces in one pound. After Renuka makes one batch o

IceJOKER [234]

16*2=32
32-.25(1.5 teaspoons) =31.75
31.75 ounces of pepper

5 0

3 years ago