Someone please help me with the answer. ASAP

Work is apreciated! If you could also say what shape this is with the formula also!

JulsSmile [24]

<h3>In geometry, a prism is a polyhedron comprising an n-sided polygon base, a second base which is a translated copy of the first, and n other faces, necessarily all parallelograms, joining corresponding sides of the two bases. All cross-sections parallel to the bases are translations of the bases. </h3>

Faces: 2+ n

Edges: 3n

Vertices: 2n

Symmetry group: Dnh,, (*n22), order 4n

Dual polyhedron: convex dual-uniform n-gonal bipyramid

Conway polyhedron notation: Pn

Rotation group: Dn,+, (n22), order 2n

<h3>formula :- The base area of a rectangular prism formula = base length x base width. The base area of a triangular prism formula = ½ x apothem length x base length. The base area of a pentagonal prism formula = 5/2 x apothem length x base length. The base area of a hexagonal prism formula = 3 x apothem length x base length.</h3>

7 0

3 years ago

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According to the graph what is the value of the constant in the equation below

Alex

It would be B (2.5)

Explanation: The Y values go up by 2.5 for each graph point higher than the one before.

8 0

3 years ago

Find a decomposition of a=⟨−5,−1,1⟩ into a vector c parallel to b=⟨−6,0,6⟩ and a vector d perpendicular to b such that c+d=a.

dezoksy [38]

The projection of vector A <em>parallel</em> to vector B is $\langle -3, 0, 3\rangle$ and the projection of vector A <em>perpendicular</em> to vector B is $\langle -2, -1, -2\rangle$ .

In this question, we need to determine all projections of a vector with respect to another vector. In this case, the projection of vector A <em>parallel</em> to vector B is defined by this formula:

$\vec a_{\parallel , \vec b} = \frac{\vec a \,\bullet\,\vec b}{\|\vec b\|^{2}}\cdot \vec b$ (1)

Where $\|\vec b\|$ is the norm of vector B.

And the projection of vector A <em>perpendicular</em> to vector B is:

$\vec a_{\perp, \vec b} = \vec a - \vec a_{\parallel, \vec b}$ (2)

If we know that $a = \langle -5, -1, 1 \rangle$ and $\vec b = \langle -6, 0, 6 \rangle$ , then the projections are now calculated:

$\vec a_{\parallel, \vec b} = \frac{(-5)\cdot (-6)+(-1)\cdot (0)+(1)\cdot (6)}{(-6)^{2}+0^{2}+6^{2}} \cdot \langle -6, 0, 6 \rangle$

$\vec a_{\parallel, \vec b} = \frac{1}{2}\cdot \langle -6, 0, 6 \rangle$

$\vec a_{\parallel, \vec b} = \langle -3, 0, 3\rangle$

$\vec a_{\perp, \vec b} = \langle -5, -1, 1 \rangle - \langle -3, 0, 3 \rangle$

$\vec a_{\perp, \vec b} = \langle -2, -1, -2\rangle$

The projection of vector A <em>parallel</em> to vector B is $\langle -3, 0, 3\rangle$ and the projection of vector A <em>perpendicular</em> to vector B is $\langle -2, -1, -2\rangle$ .

We kindly invite to check this question on projection of vectors: brainly.com/question/24160729