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tekilochka [14]
3 years ago
12

The difference of the two solutions of x^2−17x+c=0 is 1. Find the solutions.

Mathematics
1 answer:
Debora [2.8K]3 years ago
7 0
Hmm,
we see that the middle term is negative
that tells us that the factors must also be negative because if they were different signs then they would have to be 0,1, or -1,0 which wouldn't work because x doesn't factor out, well, if c=0, then it would get x=17 and 0 which doesn't work


well


what 2 numbers are 1 apart, and are negative and add to -17
-8 and -9
multily them
72

c=72
x^2-17x+72=0
if we factord
(x-8)(x-9)=0
theh zeroes are x=8 and 9
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stepladder [879]

Answer:

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General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                             \displaystyle \lim_{x \to c} x = c

Limit Rule [Variable Direct Substitution Exponential]:                                         \displaystyle \lim_{x \to c} x^n = c^n

Limit Property [Multiplied Constant]:                                                                     \displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle  \lim_{x \to 0} f(x) = 4

<u>Step 2: Solve</u>

  1. Rewrite [Limit Property - Multiplied Constant]:                                           \displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = \frac{1}{4} \lim_{x \to 0} [f(x)]^4
  2. Evaluate limit [Limit Rule - Variable Direct Substitution Exponential]:       \displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = \frac{1}{4}(4^4)
  3. Simplify:                                                                                                         \displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = 64

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

Book: College Calculus 10e

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