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Anuta_ua [19.1K]
2 years ago
6

What is the square root of -16?

Mathematics
2 answers:
liberstina [14]2 years ago
8 0
4i would be the correct answer.
Step2247 [10]2 years ago
4 0

Step-by-step explanation:

3 is the correct answer

You might be interested in
Lucia estimates the attendance at the middle school concert to be 267. The actual attendance is 304.
USPshnik [31]

Answer:

12%

Step-by-step explanation:

Divide 267 by 304

Lucia got 88% correct

Lucia got 12% Incorrect

3 0
2 years ago
Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function.
marishachu [46]

Answer:

\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18

Step-by-step explanation:

We have been given the function

f(x)=-2x^2+4x^3+3x^2+18

From the rational zeros theorem, we have

\text{Possible rational zeros}=\pm\frac{\text{Factors of constant term}}{\text{Factors of leading coefficient}}

From the given function,

Leading coefficient = 2

Factors of 2 are 1,2

Constant term = 18

Factors of constant term = 1, 2, 3, 6, 9, 18

Hence, we have

\text{Possible rational zeros}=\pm\frac{1,2,3,6,9,18}{1,2}\\\\\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18

6 0
3 years ago
9. Put these numbers in greatest to least order: 2/3, 60%, 0.06, 3/4
Iteru [2.4K]

Step-by-step explanation:

  1. 0.06
  2. 60%
  3. 2/3
  4. 3/4

hope it's right :)

8 0
3 years ago
Answer:<br> P- 56<br> P+ 56<br> p<br> 56- P
vodomira [7]
The answer is P - 56
8 0
3 years ago
Read 2 more answers
1. Consider the following hypotheses:
Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

1)

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

2)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f)  ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

7 0
3 years ago
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