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3 years ago

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The total daily cost (in dollars) of manufacturing scented candles is given by the function C(b) = 250 + 3b, where b is the numb

er of candles manufactured. Which function represents the average manufacturing cost, A, in terms of b?

1 answer:

guajiro [1.7K]3 years ago

6 0

The function that represents the average manufacturing cost in terms of b is:

A(b) = $\frac{250}{b}+3$

Total daily cost ( in dollars) of manufacturing scented candles is given by the function

C(b) = 250 + 3b

The number of candles manufactured = b

Therefore, average manufacturing cost

= Total cost / Total number of candles manufactured

For this reason,

A (b) = C(b) / b = $\frac{250+3b}{b}$ = $\frac{250}{b} +3\frac{b}{b}=\frac{250}{b}+3$

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Which solid figures have their volumes calculated using the formula V = One-thirdBh, where B is the area of the base and h is th

eimsori [14]

Answer:

A. Right Cone

C. Hexagonal prism

Step-by-step explanation:

The volume of a cone with height h and radius r is 1/3πr^2h, which is exactly one third the volume of the smallest cylinder that it fits inside.

The volume of a hexagonal pyramid with height h and radius r is 1/3πr^2h, which is exactly one third the volume of the smallest rectangular prism that it fits inside.

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3 years ago

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What is the degree of each of the following polynomial functions?

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Answer:

2,3,5,7,n

Step-by-step explanation:

Take the highest power and that is the degree of the polynomial, I don't know what the special names are

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3 years ago

Triangle E F G. Side E F is 6 meters, F G is 5 meters, E G is 7 meters. Angle F is 87 degrees and angle G is 58 degrees. Triangl

daser333 [38]

Answer:

Similar triangles

Step-by-step explanation:

Refer to attached

<u>Given:</u>

<h3>ΔEFG</h3>

EF =6 m, FG = 5m, EG = 7 m
∠F = 87°, ∠G= 58°, ∠E = 180° -(87°+58°) = 35°

<h3>ΔLJK</h3>

KL= 28 m, LJ = 24 m, JK = 20 m
∠J = 87°, ∠K = 58°, ∠L = 35°

<u>We can see that:</u>

∠F≅∠J, ∠G ≅ ∠K, ∠E ≅ ∠L

and

KL/EG = LJ/EF = JK/FG = 28/7= 24/6= 20/5 = 4

So the corresponding angles are congruent and corresponding sides are similar

We can conclude that the triangles are similar

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4 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

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2 years ago

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The area of a circle is 25pi m squared what is the circumference

IceJOKER [234]

$\bf \textit{area of a circle}\\\\
A=\pi r^2\quad 
\begin{cases}
r=radius\\
-----\\
A=25\pi 
\end{cases}\implies 25\pi =\pi r^2\implies \cfrac{25\pi }{\pi }=r^2
\\\\\\
25=r^2\implies \sqrt{25}=r\implies 5=\boxed{r}\\\\
-------------------------------\\\\
\textit{circumference of a circle}\\\\
C=2\pi r\qquad \qquad C=2\pi\left( \boxed{5} \right)\implies C=10\pi$

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3 years ago