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sukhopar [10]
3 years ago
15

What is the answer to a+5= -5a+5

Mathematics
1 answer:
9966 [12]3 years ago
7 0
the answer is a = 0
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The volume of a pyramid that fits exactly inside a cube is 18 cubic feet. what is the volume of the cube? 6 cubic feet 18 cubic
fomenos

The volume of the cube that perfectly fits an 18 ft³ pyramid is calculated as (C) 54 ft³.

<h3>What is a cube?</h3>
  • A cube is a three-dimensional solid object with six square faces, facets, or sides, three of which meet at each vertex.
  • The cube is one of the five Platonic solids and the only regular hexahedron.
  • It has six faces, twelve edges, and eight vertices.

To find the volume of the cube that perfectly fits an 18 ft³ pyramid:

We have been provided that:

  • 18 cubic feet is the volume of the pyramid.
  • Now, in order for this pyramid to fit exactly into a cube, the base of the pyramid must be square, and the height of the pyramid must be equal to the height of the cube.
  • We can conclude from this that the volume of a cube equals three times the volume of a pyramid.
  • So, the volume of the cube = 3 × 18
  • The volume of Cube = 54 ft³

Therefore, the volume of the cube that perfectly fits an 18 ft³ pyramid is calculated as (C) 54 ft³.

Know more about a cube here:

brainly.com/question/1972490

#SPJ4

The correct question is given below:

The volume of a pyramid that fits exactly inside a cube is 18 cubic feet. what is the volume of the cube?

(A) 6 cubic feet

(B) 18 cubic feet

(C) 54 cubic feet

(D) 72 cubic feet

6 0
9 months ago
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
2 years ago
Althea and her sister want to hang 5 pics on a 12-foot wall. Each pic frame must be equal distance apart from each other and the
omeli [17]

the answer to it is the frames has to be 94 ithink

8 0
3 years ago
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