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shusha [124]
3 years ago
14

Pls help, I have been on this question for legit 2 h

Mathematics
1 answer:
Nookie1986 [14]3 years ago
8 0

(4/5) 60 = 48 potatoes only, filled in correctly in the box

(1/5) 60 = 12 beets only; put that in the rightmost box

That leaves 40 for the middle, both potatoes and beets.

Answer: fill in the boxes 48 40 12

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What is the image of N for a 300° counterclockwise rotation about the center of the regular hexagon? Will give brainiest only if
OLEGan [10]

Answer:

H

Step-by-step explanation:

here, the question says that the given regular hexagon needs to be rotated counter clockwise 300°, considering the edges labels, each movement from one edge to other is 60° as 360/6 =60.

focus on N and move on anti clockwise.

When N rotates anti clockwise about center from original to position G, is 60°,

when N moves on anti clockwise about center from original to position A is 120°.

similarly, to X is 180°, to E is 240° and to H is 300°.

so, the new position of N when rotated anti clockwise about origin of the hexagon will be at H.

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3 years ago
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Answer:the first one can you pick more?

Step-by-step explanation:

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Talja [164]

Answer:

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Use the Remainder Theorem to find the remainder for (2x^3-3x^2+6)/(x-1) and state whether or not the binomial is a factor of the
Bezzdna [24]

Answer:

Remainder= 5, and the binomial (x-1) is not a factor of the given polynomial.

Step-by-step explanation:

Given polynomial is (2x^3-3x^2+6) , we have to divide this with a binomial [tex}(x-1)[/tex] using remainder theorem.

Remainder theorem says if (x-a) is a factor then remiander would be f(a)

Therefore for (x-1), \ {we find}\  f(1)}

f(1)=(2\times 1^3-3\times1^2+6)\\1^3 =1\\1^2=1\\Substituting \ this \ above\\f(1)= (2-3+6)=5

Thus the remainder is 5 and since it is not 0 , so the binomial (x-1) is not a factor of the given polynomial.

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