What is the image of point (0,3) for mapping (x,y) (x-2,y-3)

1 answer:

5 0

When it asks for the image, that means what are your new coordinates after doing the rules they give you. In this, the rule is to subtract 2 from your x value and subtract 3 from the y value. So you subtract 2 from 0 and subtract 3 from 3. This will give you the values to use for the x and y of your answer:)

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Find the constant $a$ such that\[(x^2 - 3x + 4)(2x^2 +ax + 7) = 2x^4 -11x^3 +30x^2 -41x +28.\]

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The value of constant a is -5

Further explanation:

We will use the comparison of co-efficient method for finding the value of a

So,

Given

$(x^2 - 3x + 4)(2x^2 +ax + 7)\\= x^2((2x^2 +ax + 7)-3x((2x^2 +ax + 7)+4(2x^2 +ax + 7)\\= 2x^4+ax^3+7x^2-6x^3-3ax^2-21x+8x^2+4ax+28\\Combining\ alike\ terms\\=2x^4+ax^3-6x^3+7x^2-3ax^2+8x^2-21x+4ax+28\\= 2x^4 +(a-6)x^3+(15-3a)x^2-(21-4a)x+28\\$