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Monica [59]
3 years ago
12

Someone please help me I’ll give out brainliest please dont answer if you don’t know

Mathematics
1 answer:
Anton [14]3 years ago
4 0

Answer:

see explanation

Step-by-step explanation:

j - 4 ≤ 0 ( add 4 to both sides )

j ≤ 4

The graph has a closed circle at 4 on the number line with the arrow pointing left

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Can someone help me please please?
erastovalidia [21]

Answer:

22

Step-by-step explanation:

Count the dot

6 0
2 years ago
Consider the equation 5 + x = n
Inga [223]
All of the answers will be less than five. for example if x=-3 n=2 which is less than 5
3 0
2 years ago
A biologists uses the equation l = 2a + 13.1 to estimate the length (l), in centimeters, in terms of the age (a), in months. Bas
steposvetlana [31]

Answer:

b = 3.1

Step-by-step explanation:

Since the length of one of the longer sides, a, has been given to be 6.3, plug it into the equation.

2(6.3) + b = 15.7

12.6 + b = 15.7

b = 15.7 - 12.6

3 0
3 years ago
Y = 2x <br> x + y = −6<br> the answer should be like this (x,y) or ( , )
melisa1 [442]

Answer:

(-2, -4)

Step-by-step explanation:

Given:

The system of equations to solve is given as:

y=2x\\\\x+y=-6

In order to solve this system of linear equations, we use substitution method.

In substitution method, we substitute the value of any one variable in the other equation.

So, we substitute the value of 'y' from first equation in the second equation.

This gives,

x+2x=-6\\\\3x=-6

Dividing both sides by 3, we get:

\frac{3x}{3}=\frac{-6}{3}\\\\x=-2

Now, y=2x=2\times -2=-4

Therefore, the solution is (-2, -4)

6 0
3 years ago
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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