Answer:
im sorry but i think this question is higher than 5 points
(-x²-2x+6)+(-9x²-7x+6) = -x²-2x+6-9x²-7x+6 = -10x² - 9x + 12
(X^2 -5x+6)(x-5) = x^3- 10x^2 +31x -30
Since we have two points (2,54) and (4,54), we can assume a linear function and solve for the slope and intercept. So if V=kt+b where t is time passed and v is velocity of the car, we can plug in and solve for k and t, that would give t as a function of v, and you can graph it
The correct answer is: [B]: "Difference of Cubes".
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Explanation:
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Note that the equation/identity for the "difference of cubes" is expressed as:
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" a</span>³ − b³ = (a − b)(a² + ab + b²) " ;
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Note the given equation: " 19 = 27 </span>− 8 " ; → (which is true).
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The "right hand side" of this equation:
→ " 27 − 8 " ; contains two numbers:
→ "27" and "8" ; both of which are "cubes" ;
→ that is: ∛27 = 3 ; ↔ 3³ = 3 * 3 * 3 = 9 * 3 = 27 ; <u><em>and</em></u>:
∛ 8 = 2 ; ↔ 2³ = 2 * 2 * 2 = 4 * 2 = 8 ;
→ <u>AND:</u> "8" is being <u>SUBTRACTED</u> from "27" ;
→ (hence, the "difference of squares" polynomial identity);
So: given: " 19 = 27 − 8 " ;
→ Rewrite as:
" 19 = 3³ − 2³ " ;
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Now, consider the identity equation for the "difference of squares":
→ " a³ − b³ = (a − b)(a² + ab + b²) " ;
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Take: " 19 = 3³ − 2³ " ;
and rewrite as:
→ 3³ − 2³ = 19 ;
So: (a³ − b³) = 3³ − 2³ ;
a = 3 ; b = 2 ;
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Plug in these values:
" a³ − b³ = (a − b)(a² + ab + b²) " ;
→ 3³ − 2³ ≟ [3 − 2) [ 3² + (3*2) + 2² ] ≟ 19 ? ;
→ 27 − 8 ≟ (1) (9 + 6 + 4) ≟ 19 ? ;
→ 19 ≟ (1) (15 + 4) ≟ 19 ? ;
→ 19 ≟ (1) (19) ≟ 19 ? ;
→ 19 ≟ 19 ≟ 19 ? ;
→ 19 = 19 = 19 ! Yes!
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