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4 years ago

13

Isabella is having going to the movies with some friends each friend get 6/8 cups of popcorn if there are 7 friends at the movie

s how much popcorn did the friends get

1 answer:

aev [14]4 years ago

8 0

5.25 cups of popcorn.

You will get this by multiplying the total number of friends by how much popcorn each of them is getting.

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Can some one answer this?

Anvisha [2.4K]

Answer:

1. They ate 1/6 of the donuts

2. 2/3

Step-by-step explanation:

1. They bought 18 donuts. 1/2 of 18 is 9. 1/3 of 9 is 3. To find the fraction, divide 3 by 18. Simplify the fraction. You are left with 1/6.

2. 1 1/3=4/3×1/2=4/6=2/3

5 0

2 years ago

A farmer has a total of 23 pigs and chickens. The 23 animals have a total

andre [41]

Answer:

7 pigs, 16 chickens

Step-by-step explanation:

p= pigs, c= chickens, legs= 60, number of pigs & chickens= 23

System of Equations:

Equation 1: $p+c=23$ would be $23-p=c$
Equation 2: $4p+2c=60$ for the legs

Then, use substitution & solve:

$4p+2(23-p)=54\\4p+46-2p=54\\4p-2p=8\\2p=8\\p=7$

Now, plug $p=7$ back into $23-p=c$ to get c

$23-p=c\\23-7=c\\c=16$

Answer: 7 pigs, 16 chickens or $p=7$ and $c=16$

To check your work, plug those solutions back into the system.

<em>------</em>

<em>(I hope that helps & good luck! <3)</em>

3 0

2 years ago

Monica is planning a trip to Europe for vacation. She wants to pack appropriately but she doesn't understand the average tempera

Marrrta [24]

Answer:

53.6

Step-by-step explanation:

$C=\frac{5}{9}(F-32) \\ \\ 12=\frac{5}{9}(F-32) \\ \\ 21.6=F-32 \\ \\ F=53.6$

7 0

2 years ago

-37= n/23<br> what is n?

Vlada [557]

-37 = <u>n
</u> 23
<u>
</u>to solve for n, undo what is happening to the n. if the n is being divided by 23 to get -37, then multiply both sides by 23 to figure out what n is.
<u>
</u>-37 * 23 = <u>n * 23
</u> 23
<u>
</u>-851 = n

3 0

3 years ago

Read 2 more answers

All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

6 0

3 years ago