Question

rewrite the function in standard form (ax + by = c) $y = \frac{3}{8}x + 40$
recall - 2 steps (eliminate the fraction and then isolate the constant value)
-3x + y = 5
-3x + 8y = 5
3x + y = 320
3x + 8y = 40
-3x + 8y = 320
-3x + y = 40
21 numeric 1 point
find f(-5) for $x^5 + 3x^4 - 4x^3 + 12x^2 - 2x - 2$
answer

Explanation:

First Problem (Rewrite to Standard Form)

Step1: Eliminate the fraction

Multiply all terms by 8:
$$8y = 3x + 320$$

Step2: Isolate the constant

Subtract $3x$ from both sides:
$$-3x + 8y = 320$$

Step1: Substitute $x=-5$ into the function

$$f(-5) = (-5)^5 + 3(-5)^4 - 4(-5)^3 + 12(-5)^2 - 2(-5) - 2$$

Step2: Calculate each term

$$(-5)^5=-3125,\ 3(-5)^4=3(625)=1875,\ -4(-5)^3=-4(-125)=500,\ 12(-5)^2=12(25)=300,\ -2(-5)=10$$

Step3: Sum all terms

$$f(-5) = -3125 + 1875 + 500 + 300 + 10 - 2$$

Step4: Compute the final total

$$f(-5) = (-3125) + 1875 = -1250;\ -1250 + 500 = -750;\ -750 + 300 = -450;\ -450 + 10 = -440;\ -440 - 2 = -442$$

Answer:

-3x + 8y = 320

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Second Problem (Find $f(-5)$)