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Lubov Fominskaja [6]
2 years ago
5

Find the number of faces, edges, and vertices of this shape.​

Mathematics
2 answers:
aliya0001 [1]2 years ago
4 0
Faces:7

Vertices:12

Edges:7
sattari [20]2 years ago
3 0
Faces: 7

Edges: 12

Vertices: 7
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How much is one can of peaches if the price for three is 2.99?
BabaBlast [244]
If the price for three of them is $2.99, then you divide that by three to find the price for one can. This would equal .9966666667, which could probably be rounded to 99 cents or just a dollar per can. Hope that helps.
7 0
3 years ago
Simplify to create an equivalent expression. 2(3r+7)−(2+r)
tamaranim1 [39]

2(3r+7)−(2+r)

Use distributive property:

6r + 14 - 2 - r

Simplify by combining like terms:

5r +12

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3 years ago
Which sign makes the statement true?<br> &gt;<br> =
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Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
LCM of x&lt;sup&gt;2&lt;/sup&gt;+5x+6 and x&lt;sup&gt;2&lt;/sup&gt;-x-6 is ………………………
sergejj [24]

Answer:

(x^2 - 9)(x + 2)

Step-by-step explanation:

Given:

x^2 + 5x + 6

x^2 - x - 6

Required:

LCM of the polynomials

SOLUTION:

Step 1: Factorise each polynomial

x^2 + 5x + 6

x^2 + 3x + 2x + 6

(x^2 + 3x) + (2x + 6)

x(x + 3) + 2(x + 3)

(x + 2)(x + 3)

x^2 - x - 6

x^2 - 3x +2x - 6

x(x - 3) + 2(x - 3)

(x + 2)(x - 3)

Step 2: find the product of each factor that is common in both polynomials.

We have the following,

x^2 + 5x + 6 = (x + 2)(x + 3)

x^2 - x - 6 = (x + 2)(x - 3)

The common factors would be: =>

(x + 2) (this is common in both polynomials, so we would take just one of them as a factor.

(x + 3) and,

(x - 3)

Their product = (x - 3)(x + 3)(x +2) = (x^2 - 9)(x + 2)

4 0
3 years ago
Determine whether AB ← → and CD ← → − are parallel, perpendicular, or neither. A(-6, 2), B(−3, -4), C(1, -3), D(3, 4)
guapka [62]
Neither they connect but don’t make a 90 degree angle

6 0
2 years ago
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