All categories

2 years ago

-4(-11+4n) -3( -2n+9)

Mathematics

2 answers:

Ludmilka [50]2 years ago

6 0

Answer:

Step-by-step explanation:

belka [17]2 years ago

3 0

Answer:17-10n

Step-by-step explanation:

You might be interested in

Find sec θ if <br> tan θ = 15/8<br> and θ terminates in QIII.

EastWind [94]

Tangent = opposite/adjacent
tangent = 15 / 8
If opp = 15 and adj = 8 then
hypotenuse^2 = 15^2 + 8^2
hypotenuse^2 = 225 + 64
hypotenuse^2 = 289
hypotenuse = 17

secant = hypotenuse / adjacent
secant = 17 / 8
secant = 2.125

5 0

3 years ago

If a cooler contains six cola, two grape, two orange, and four lemon-lime sodas, what is the probability of selecting a cola, gi

frutty [35]

Answer:

ok so first of all there is 14 sodas in the cooler(6+2+2+4)

and 6 of these 14 are cola 6/14 or 3/7

so 3/7 then to give your self one we multiple it by 3/7

3/7*3/7=0.18367346938

we multiple by 10

18.367346938

so the probility is 18.37%

Hope This Helps!!!

6 0

2 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

Carl and Tamera want to compare their tests in math class. They have taken six tests. Should they use the mean, median or mode t

Roman55 [17]

Answer:

Mean

Step-by-step explanation:

I said mean because it would allow both Carl and Tamera to average their scores and compare one side by side rather than comparing 6 each which would just be time consuming and it seems like it wouldn't be any other choices.

7 0

2 years ago

Mariah just read the following passage in a short story:

Nana76 [90]

Answer:

D. enlargement

Step-by-step explanation:

enlargement means to increase something

3 0

2 years ago

Read 2 more answers