QUESTION IMAGE
Question
complete the tasks to subtract the polynomials vertically.
(1.3t³ + 0.4t² - 24t) - (0.6t² + 8 - 18t)
what is the additive inverse of the polynomial being subtracted?
-0.6t² + (-8) + (-18t)
-0.6t² + (-8) + 18t
-0.6t² + 8 - 18t
0.6t² + (-8) + 18t
complete
choose the correct values for a, b, c, and d that align like terms to find the sum vertically.
1.3t³ + 0.4t² + (-24t)
+ a + b + c + d
a = (-0.6t²) b = (-8) c = 18t d = 0t³
a = 0t³ b = (-0.6t²) c = 18t d = (-8)
a = 0t³ b = (-0.6t²) c = (-8) d = 18t
complete
what is the difference of the polynomials?
0.7t³ - 0.2t² - 6t - 8
0.7t³ - t² - 6t - 8
1.3t³ - t² + 6t - 8
1.3t³ - 0.2t² - 6t - 8
done
For the first part (additive inverse of the subtracted polynomial):
The polynomial being subtracted is \(0.6t^{2}+8 - 18t\). The additive inverse of a polynomial \(P(t)\) is \(-P(t)\). So we distribute the negative sign: \(-(0.6t^{2}+8 - 18t)=- 0.6t^{2}-8 + 18t=-0.6t^{2}+(-8)+18t\).
Step 1: Write the polynomials vertically with like terms aligned
We have \(1.3t^{3}+0.4t^{2}-24t\) and we are subtracting \(0.6t^{2}+8 - 18t\). When we write the subtraction as addition of the additive inverse, we have:
\[
\]
Step 2: Add the like terms
- For the \(t^{3}\) terms: \(1.3t^{3}+0t^{3}=1.3t^{3}\)
- For the \(t^{2}\) terms: \(0.4t^{2}+(-0.6t^{2})=(0.4 - 0.6)t^{2}=-0.2t^{2}\)
- For the \(t\) terms: \(-24t + 18t=(-24 + 18)t=-6t\)
- For the constant terms: \(0+(-8)=-8\)
So the result is \(1.3t^{3}-0.2t^{2}-6t - 8\)
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\(-0.6t^{2}+(-8)+18t\) (the second option in the first multiple - choice question)