Question 3 (2 points)Using the statement below, determine the hypothesis and conclusion,

If two integers are even, then their sum is even

Column A

Column B

1.

Hypothesis

a. If two integers are even

2.

Conclusion

b. then their sum is even

c. two integers are even

d. their sum is even

Tanya has 26 quarters and 18 dimes.

Given that Tanya has 44 quarters and dimes which worth $8.30, we need to find the number of each coin type,

To find the same we will use the concept of system of Linear equations,

Let's solve this problem step by step.

Let's assume Tanya has x quarters and y dimes.

The value of x quarters is 25x cents.

The value of y dimes is 10y cents.

According to the given information, the total value of the quarters and dimes is $8.30, which is equivalent to 830 cents. So we have the equation:

25x + 10y = 830 ...........(Equation 1)

Tanya has 44 coins in total:

x + y = 44 ...........(Equation 2)

Now, we can solve this system of equations (Equation 1 and Equation 2) to find the values of x and y.

Multiplying Equation 2 by 25, we get:

25x + 25y = 1100 ...........(Equation 3)

Subtracting Equation 3 from Equation 1, we eliminate the x term:

25x + 10y - (25x + 25y) = 830 - 1100

-15y = -270

Dividing both sides by -15, we get:

y = (-270)/(-15) = 18

Substituting the value of y back into Equation 2, we can find x:

x + 18 = 44

x = 44 - 18 = 26

Therefore, Tanya has 26 quarters and 18 dimes.

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Answer:

.06 meters

Step-by-step explanation:

3/5=.6

.6/9=.06

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.