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Romashka-Z-Leto [24]

2 years ago

14

The bottom and top of a box are rectangles twice as long as they are wide. Find the volume of the box if it is four feet high an

d has a total surface area of 220 feet squared.

1 answer:

Sav [38]2 years ago

5 0

Let x = the width

then

2x = the length

:

The box dimensions: 2x by x by 4

Given the surface area:

2(2x*x) + 2(2x*4) + 2(x*4) = 220

:

4x^2 + 16x + 8x = 220

A quadratic equation:

4x^2 + 24x - 220 = 0

simplify, divide by 4

x^2 + 6x - 55 = 0

Factor

(x+11)(x-5) = 0

The positive solution is what we want here:

x = 5 ft is the width

then

2(5) = 10 ft is the length

:

Find the volume

10 * 5 * 4 = 200 cu/ft is the volume

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Mark bought refreshments. Sandwiches were 2.50 each and bottles of water were $1.25 each. He spent $356.25 on a total of 180 ite

Over [174]

First you have to write the equation. in the scenario, use standard form.
Ax+By=C
plug the numbers in. A=2.50, B=1.25, and C is the total, 356.25. the 180 doesn't come in quite yet.
your equation is 2.50x+1.25y=356.25. now, since they only bought 180 items, you can't go past that.

I am sorry, but I am about to leave for school, and therefore do not have enough time to answer the last of your question. I hope the part I could answer has helped you.

7 0

3 years ago

For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:

Oksi-84 [34.3K]

The half-range sine series is the expansion for $f(t)$ with the assumption that $f(t)$ is considered to be an odd function over its full range, $-1$ . So for (a), you're essentially finding the full range expansion of the function

$f(t)=\begin{cases}2-t&\text{for }0\le t$

with period 2 so that $f(t)=f(t+2n)$ for $|t|$ and integers $n$ .

Now, since $f(t)$ is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

$f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L$

where

$b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt$

In this case, $L=1$ , so

$b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt$
$b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{4-2(-1)^n}{n\pi}$

The half-range sine series expansion for $f(t)$ is then

$f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t$

which can be further simplified by considering the even/odd cases of $n$ , but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming $f(t)$ is even/symmetric across its full range. In other words, you are finding the full range series expansion for

$f(t)=\begin{cases}2-t&\text{for }0\le t$

Now the sine series expansion vanishes, leaving you with

$f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L$

where

$a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt$

for $n\ge0$ . Again, $L=1$ . You should find that

$a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3$

$a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt$
$a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}$
$a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}$

Here, splitting into even/odd cases actually reduces this further. Notice that when $n$ is even, the expression above simplifies to

$a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0$

while for odd $n$ , you have

$a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}$

So the half-range cosine series expansion would be

$f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t$

Attached are plots of the first few terms of each series overlaid onto plots of $f(t)$ . In the half-range sine series (right), I use $n=10$ terms, and in the half-range cosine series (left), I use $k=2$ or $n=2(2)-1=3$ terms. (It's a bit more difficult to distinguish $f(t)$ from the latter because the cosine series converges so much faster.)

5 0

3 years ago

the length of the sides of a square are initially 0 cm and increase at a constant rate of 11 cm per second. suppose the function

Likurg_2 [28]

The function of the area of the square is A(t)=121 $t^{2}$

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about The function of the area of the square is A(t)=121 $t^{2}$

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about area here:

brainly.com/question/27683633

#SPJ4

3 0

1 year ago

jason spends 3.24 on 6 chocolate chip cookies. The cookies all cost the same amount. What is the cost of each cookie

Vlada [557]

Divide 3.24 with 6 and the answer will be found

$0.54

7 0

2 years ago

Read 2 more answers

Without multiplying, tell which product is larger and why.

Mademuasel [1]

Answer:

because 6 is larger then 4

Step-by-step explanation:

up

5 0

2 years ago