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2 years ago

Martin wants to ship a souvenir baseball bat that is 16 inches long in this box.

Mathematics

1 answer:

stich3 [128]2 years ago

6 0

Answer:

Step-by-step explanation:

Remark

The first job is to find the diagonal along the bottom. Pretend that the sides and top do not exist for just a moment.

Givens

It becomes a Pythagorean problem.

a = 12

b = 9

c = ?

Solution

a^2 + b^2 = c^2

144 + 81 = c^2

c^2 = 225

sqrt(c^2) = sqrt(225)

c = 15

Remark

Now you look at the diagram again, a little more closely this time.

The diagonal that you want is from the lower right to the upper closest left.

c = 15

d = 8

e = ?

e^2 = c^2 + d^2

e^2 = 15^2 + 8^2

e^2 = 225 + 64

e^2 = 289

e = 17

The shipping box is large enough

Note

There is one think you can note

You can do this entire problem in one step

a = 12 b = 9 c= 8

a^2 + b^2 + c^2 = diagonal from lower right to upper left^2

12^2 + 9^2 + 8^2 = diagonal^2

144 + 81 + 64 = diagonal ^2

289 = diagonal^2

diagonal = 17 just as it did before.

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Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

3 0

3 years ago

NEED HELP ASAP PLEASE

Ilya [14]

Answer:

see explanation

Step-by-step explanation:

Given

A = $\frac{1}{2}$ bh