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Rus_ich [418]
3 years ago
15

GUYS PLZ HELP ME!!!!!!

Mathematics
1 answer:
siniylev [52]3 years ago
7 0

Answer:

B E F G

Step-by-step explanation:

Polygons are 2 dimensional shapes with straight edges (no curved edges). The shape needs to be closed up entirely to be a polygon. It cannot have any intersecting line segments

A can't be a polygon because the shape isn't closed up entirely

C can't be a polygon because it is a 3d shape

D can't be a polygon because it has curved edges.

H can't be a polygon because it isn't closed up (there are no line segments).

I can't be a polygon because polygons cannot have intersecting lines

J can't be a polygon because polygons cannot have intersecting lines

I hope this helps! If it does, a rate or brainliest would be much appreciated. :)

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[MA5T10 Dj Name the property being illustrated.
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Answer:

c

Step-by-step explanation:

cause if you do both it will be the same answer with is also called commutatative

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3 years ago
Find the set of values of k for which the line y=kx-4 intersects the curve y=x²-2x at 2 distinct points?
Sonbull [250]

Answer:

-6 < k < 2

Step-by-step explanation:

Given

y = x^2 - 2x

y =kx -4

Required

Possible values of k

The general quadratic equation is:

ax^2 + bx + c = 0

Subtract y = x^2 - 2x and y =kx -4

y - y = x^2 - 2x - kx +4

0 = x^2 - 2x - kx +4

Factorize:

0 = x^2 +x(-2 - k) +4

Rewrite as:

x^2 +x(-2 - k) +4=0

Compare the above equation to: ax^2 + bx + c = 0

a = 1

b= -2-k

c =4

For the equation to have two distinct solution, the following must be true:

b^2 - 4ac > 0

So, we have:

(-2-k)^2 -4*1*4>0

(-2-k)^2 -16>0

Expand

4 +4k+k^2-16>0

Rewrite as:

k^2 + 4k - 16 + 4 >0

k^2 + 4k - 12 >0

Expand

k^2 + 6k-2k - 12 >0

Factorize

k(k + 6)-2(k + 6) >0

Factor out k + 6

(k -2)(k + 6) >0

Split:

k -2 > 0 or k + 6> 0

So:

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To make the above inequality true, we set:

k < 2 or k >-6

So, the set of values of k is:

-6 < k < 2

3 0
3 years ago
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saul85 [17]

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