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Gennadij [26K]
3 years ago
13

Hey lol pls help <3

Mathematics
1 answer:
Anika [276]3 years ago
8 0
A. 80.200 random letters susboahajsgsjahaj
You might be interested in
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
3.
natima [27]

Answer:

See explanation

Step-by-step explanation:

Let x be the number of spade shovels, y -the number of flat shovels and z - the number of square showels sold that day.

The store keeps an inventory of 80 shovels, then

x+y+z=80

The store always buy twice as many spade shovels as square, so

x=2z

The total cost of all shovels is

16x+9.60y+12.80z=1,072

a) The system of three equations is

\left\{\begin{array}{l}x+y+z=80\\ \\x=2z\\ \\16x+9.60y+12.80z=1,072\end{array}\right.

b) In matrix form this is

\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}80\\0\\1,072\end{array}\right)

c) The determinant is

\left\|\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right\|=0-32+9.60-0+19.20-12.80=-16

d) Find three determinants:

\left\|\begin{array}{ccc}80&1&1\\ 0&0&-2\\ 1,072&9.60&12.80\end{array}\right\|=0-2,144+0-0+1,536-0=-608

\left\|\begin{array}{ccc}1&80&1\\ 1&0&-2\\ 16&1,072&12.80\end{array}\right\|=0-2,560+1,072-0+2,144-1,024=-368

\left\|\begin{array}{ccc}1&1&80\\ 1&0&0\\ 16&9.60&1,072\end{array}\right\|=0+0+768-0-0-1,072=-304

So,

x=\dfrac{-608}{-16}=38\\ \\y=\dfrac{-368}{-16}=23\\ \\z=\dfrac{-304}{-16}=19

e) If the store doubled all prices and inventory, then the new matrix is

\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 32&19.20&25.60\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}160\\0\\2,144\end{array}\right)

7 0
3 years ago
An athletic field is a 60 yd60 yd​-by-120 yd120 yd ​rectangle, with a semicircle at each of the short sides. A running track 101
gregori [183]

Distance of each track are:

D₁ = 428.5 yd

D₂ = 436.35 yd

D₃ = 444.20 yd

D₄ = 452.05 yd

D₅ = 459.91 yd

D₆ = 467.76 yd

D₇ = 475.61 yd

D₈ = 483.47 yd

<u>Explanation:</u>

Given:

Track is divided into 8 lanes.

The length around each track is the two lengths of the rectangle plus the two lengths of the semi-circle with varying diameters.

Thus,

D = 2(120) + 2. \frac{1}{2} \pi d\\\\D = 240 + \pi d

Starting from the innermost edge with a diameter of 60yd.

Each lane is 10/8 = 1.25yd

So, the diameter increases by 2(1.25) = 2.5 yd each lane going outward.

So, the distances are:

D₁ = 240 + π (60) → 428.5yd

D₂ = 240 + π(60 + 2.5) → 436.35 yd

D₃ = 240 + π(60 + 5) → 444.20 yd

D₄ = 240 + π(60 + 7.5) → 452.05 yd

D₅ = 240 + π(60 + 10) → 459.91 yd

D₆ = 24 + π(60 + 12.5) → 467.76 yd

D₇ = 240 + π(60 + 15) → 475.61 yd

D₈ = 240 + π(60 + 17.5) → 483.47 yd

4 0
4 years ago
Three points not in a straight line make up a triangle. Lyle asks, “Where are the locations I should put a fourth point to make
Archy [21]
Connect them in a traingle
put a point ANYWHERE outside the triangle
connect them
tada
8 0
4 years ago
The graph shows the number of miles you and a friend run each week while training for a race.
vlada-n [284]

Answer:

45

Step-by-step explanation:

4 0
3 years ago
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