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hichkok12 [17]
3 years ago
8

Judy solved the quadratic equation

Mathematics
2 answers:
cestrela7 [59]3 years ago
8 0

Answer:

A. Judy made a mistake between Steps 1 and 2.

She did not factor x² – 16=0 correctly.

x² – 16=0 is factored as:

(x +4) (x-4) = 0

Therefore, x = -4 and x = 4

Step-by-step explanation:

Lapatulllka [165]3 years ago
5 0

Answer:

I think it's D

Step-by-step explanation:

Because all of them are true

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What is the answer and how do you solve <br><br><br> 21x^3/3x^−1
Sunny_sXe [5.5K]

Answer:

  as written: 7x^2

  perhaps intended: 7x^4

Step-by-step explanation:

As written, the Order of Operations specifies that the only denominator factor is 3, so the expression is interpreted as ...

  \dfrac{21x^3}{3}x^{-1}=\dfrac{21}{3}x^{3-1}=7x^2

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Sometimes grouping of denominator factors is intended, but parentheses are not written as they should be. If the intention is ...

  21x^3/(3x^-1)

then the simplification is ...

  \dfrac{21x^3}{3x^{-1}}=\dfrac{21}{3}x^{3-(-1)}=7x^4

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The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

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Mandarinka [93]

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6.42 x 10^-8

Step-by-step explanation:

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Use truth tables to show that the following statements are logically equivalent. ∼ P ⇔ Q = (P ⇒∼ Q)∧(∼ Q ⇒ P)
Alina [70]

Answer:  The given logical equivalence is proved below.

Step-by-step explanation:  We are given to use truth tables to show the following logical equivalence :

∼ P ⇔ Q ≡ (P ⇒∼ Q)∧(∼ Q ⇒ P)

We know that

two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.

The truth table is as follows :

P     Q     ∼ P     ∼Q   ∼ P⇔ Q    P ⇒∼ Q    ∼ Q ⇒ P     (P ⇒∼ Q)∧(∼ Q ⇒ P)

T     T         F        F             F            F                   T                       F

T     F         F        T             T             T                   T                       T

F     T         T        F             T            T                   T                       T

F     F         T        T             F            T                   F                       F

Since the corresponding truth vales for ∼ P ⇔ Q and (P ⇒∼ Q)∧(∼ Q ⇒ P) are same, so the given propositions are logically equivalent.

Thus, ∼ P ⇔ Q ≡ (P ⇒∼ Q)∧(∼ Q ⇒ P).

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Step-by-step explanation:

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