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2 years ago

Which of the competitors is closest to the checkpoint? 0 is the checkpoint btw

Mathematics

1 answer:

dusya [7]2 years ago

7 0

Florence I'm pretty sure

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For each of the following, use logic laws to decide whether the statement is a tautology contradiction, or neither. a (a) (A B)

spayn [35]

Answer:

The statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ is a contingency.

The statement $(P\rightarrow \lnot P)\land P$ is a contradiction.

Step-by-step explanation:

A tautology is a proposition that is always true.

A contradiction is a proposition that is always false.

A contingency is a proposition that is neither a tautology nor a contradiction.

a) To classify the statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ , you need to use the logic laws as follows:

$(A\rightarrow \lnot B)\land (B\rightarrow A) \equiv$

$\equiv (\lnot A \lor\lnot B)\land(\lnot B \lor A)$ by the logical equivalence involving conditional statement.

$\equiv (\lnot B\lor \lnot A )\land(\lnot B \lor A)$ by the Commutative law.

$\equiv \lnot B \lor (\lnot A \land A)$ by Distributive law.

$\equiv \lnot B \lor (A \land \lnot A)$ by the Commutative law.

$\equiv \lnot B \lor F$ by the Negation law.

Therefore the statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ is a contingency.

b) To classify the statement $(P\rightarrow \lnot P)\land P$ , you need to use the logic laws as follows:

$(P\rightarrow \lnot P)\land P \equiv$

$\equiv (\lnot P \lor \lnot P)\land P$ by the logical equivalence involving conditional statement.

$\equiv P \land (\lnot P \lor \lnot P)$ by the Commutative law.

$\equiv (P \land \lnot P) \lor (P \land \lnot P)$ by Distributive law.

$\equiv F \lor F \equiv F$ by the Negation law.

Therefore the statement $(P\rightarrow \lnot P)\land P$ is a contradiction.

8 0

2 years ago

Can someone pls describe how to do this.. and thank you

luda_lava [24]

Mathematics includes the study of such topics as quantity, structure, space, and change. It has no generally accepted definition. Mathematicians seek and use patterns to formulate new conjectures; they resolve the truth or falsity of such by mathematical proof.

Step-by-step explanation:

Mathematics is the study of numbers, shapes and patterns. The word comes from the Greek word "μάθημα" (máthema), meaning "science, knowledge, or learning", and is sometimes shortened to maths (in England, Australia, Ireland, and New Zealand) or math (in the United States and Canada). ... Numbers: how things can be counted.

Therefore, Mathematics is the abstract study of topics such as quantity (numbers), structure, space, and change!

7 0

2 years ago

Read 2 more answers

Let 3.27<<5.89 represent an interval on the number line. Find the value that is in the middle of the interval.

laila [671]

Answer:

The middle value of the interval is 4.58

Step-by-step explanation:

Consider the provided interval.

We need to find the value that is in the middle of the interval.

We can find the middle value of the interval by adding the upper and lower limit and divide the sum of the upper and lower limits by 2.

Here the upper limit is 3.27 and lower limit is 5.89.

$\frac{3.27+5.89}{2}=\frac{9.16}{2}=4.58$

Hence, the middle value of the interval is 4.58

4 0

2 years ago

4x + 6 - x = 7 + 2 x

rodikova [14]

Answer:

x = 1

Step-by-step explanation:

4x + 6 - x = 7 + 2x

3x + 6 = 7 + 2x

3x - 2x = 7 - 6

x = 1

8 0

2 years ago

Read 2 more answers

An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t s

Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, $s=120\ ft/s$

The height 'h' of the debris above the ground is given as:

$h(t)=-16t^2+120t$

As per question, $h(t)=56\ ft$ . Therefore,

$56=-16t^2+120t$

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

$-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0$

Using quadratic formula to solve for 't', we get:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s$

Therefore, the debris will reach a height of 56 ft twice.

When time $t=0.5\ s$ during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at $t=7\ s$ , the height is 56 ft.

5 0

2 years ago