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3 years ago

7

a cereal box has dimensions of 2 in., 5 1/3 in., and 10 3/4 in. If the box contains 8 servings, how much volume does each servin

g take up?

2 answers:

zaharov [31]3 years ago

5 0

Answer:

Step-by-step explanation:

Volume is found by length multiplied by width multiplied by height. Therefore our first equation is 2 multiplied by 5 and 1/3 or 16/5 as a mixed number multiplied by 10 3/4 or 43/10 as a mixed number. This will give us the total volume. As there are 8 servings, we will need to divide by 8 or multiply by 1/8 to find the volume of each serving. The final answer is 86/25 which simplified to 3 and 11/25

Trava [24]3 years ago

4 0

Answer:

"Each serving occupies a volume of 14.33 in³ in the box of cereals."

Step-by-step explanation:

It is Given That :

A cereal box has dimensions of 2 in., 5 1/3 in., 10 3/4 in.

(Volume = length * width * height)

Firstly, we convert each value into improper fractions.

So it'll be :

2 in. → 2/1 inches

5 1/3 in → 16/3 inches

10 3/4 in → 43/4 inches

A) Volume = 2/1 in * 16/3 in * 43/4 in = 2*16*43 / 1*3*4 = 1,376 in³ / 12 = 114.67 in³

B) 8 servings

114.67 in³ / 8 servings = 14.33 in³ per serving

So your answer will be : Each serving occupies a volume of 14.33 in³ in the box of cereals.

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A pyramid has a square base of length 8cm and a total surface area of 144cm². Find the volume of the pyramid. (Please use Pythag

Sveta_85 [38]

Answer:

$\displaystyle V_{ \text{pyramid}}= 64 \: {cm}^{3}$

Step-by-step explanation:

we are given surface area and the length of the square base

we want to figure out the Volume

to do so

we need to figure out slant length first

recall the formula of surface area

$\displaystyle A_{\text{surface}}=B+\dfrac{1}{2}\times P \times s$

where B stands for Base area

and P for Base Parimeter

so

$\sf\displaystyle \: 144=(8 \times 8)+\dfrac{1}{2}\times (8 \times 4) \times s$

now we need our algebraic skills to figure out s

simplify parentheses:

$\sf\displaystyle \: 64+\dfrac{1}{2}\times32\times s = 144$

reduce fraction:

$\sf\displaystyle \: 64+\dfrac{1}{ \cancel{ \: 2}}\times \cancel{32} \: ^{16} \times s = 144 \\ 64 + 16 \times s = 144$

simplify multiplication:

$\displaystyle \: 16s + 64 = 144$

cancel 64 from both sides;

$\displaystyle \: 16s = 80$

divide both sides by 16:

$\displaystyle \: \therefore \: s = 5$

now we'll use Pythagoras theorem to figure out height

according to the theorem

$\displaystyle \: {h}^{2} + (\frac{l}{2} {)}^{2} = {s}^{2}$

substitute the value of l and s:

$\displaystyle \: {h}^{2} + (\frac{8}{2} {)}^{2} = {5}^{2}$

simplify parentheses:

$\displaystyle \: {h}^{2} + (4 {)}^{2} = {5}^{2}$

simplify squares:

$\displaystyle \: {h}^{2} + 16 = 25$

cancel 16 from both sides:

$\displaystyle \: {h}^{2} = 9$

square root both sides:

$\displaystyle \: \therefore \: {h}^{} = 3$

recall the formula of a square pyramid

$\displaystyle V_{pyramid}=\dfrac{1}{3}\times A\times h$

where A stands for Base area (l²)

substitute the value of h and l:

$\sf\displaystyle V_{ \text{pyramid}}=\dfrac{1}{3}\times \{8 \times 8 \}\times 3$

simplify multiplication:

$\sf\displaystyle V_{ \text{pyramid}}=\dfrac{1}{3}\times 64\times 3$

reduce fraction:

$\sf\displaystyle V_{ \text{pyramid}}=\dfrac{1}{ \cancel{ 3 \: }}\times 64\times \cancel{ \: 3}$

hence,

$\sf\displaystyle V_{ \text{pyramid}}= 64 \: {cm}^{3}$

8 0

2 years ago

At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

1.

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

2.

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

3.

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

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10 months ago

1) Graph y = f(x) = 3(2)%. Adjust the equation to make the y-intercept (0, -3). Which response is not correct?

Karolina [17]

Using the concept of y-intercept, it is found that the response that is not correct is given by:

B) Replace x with -x.

<h3>What is the y-intercept of a function?</h3>

It is the value of the function when x = 0, hence it is the point (0, f(0)).

In this problem, the function is:

$f(x) = 3(2)^x$

The y-intercept is given by:

$f(0) = 3(2)^0 = 3$

If we just replace x with -x, we still have the same expression as above, hence the y-intercept does not change and option B is the answer to this question.

More can be learned about y-intercepts at brainly.com/question/24737967

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2 years ago

An ant starts at (0, 0), and only makes moves of length 1 in the positive x directions or the positive y direction. How many pat

AlladinOne [14]

Answer:

The number of paths that are there from the ant that end at (3, 3) but never pass through (2, 3) is 4.

Step-by-step explanation:

The possible pathways are as follows:

(0, 0) → (1, 0) → (2, 0) → (3, 0) → (3, 1) → (3, 2) → (3, 3)

(0, 0) → (1, 0) → (2, 0) → (2, 1) → (2, 2) → (3, 2) → (3, 3)

(0, 0) → (1, 0) → (1, 1) → (1, 2) → (2, 2) → (3, 2) → (3, 3)

(0, 0) → (0, 1) → (0, 2) → (1, 2) → (2, 2) → (3, 2) → (3, 3)

Thus, the number of paths that are there from the ant that end at (3, 3) but never pass through (2, 3) is 4.

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3 years ago

Convert 60°F to degrees Celsius. If necessary, round your answer to the nearest tenth of a degree.

Luba_88 [7]

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