MATHEMATICS HIGH SCHOOL

If p is true and q is false, the p -> q is always, sometimes, never true.When p is false and q is true, then p or q is always, sometimes, never true.

If p is true and ~ q is false, then p -> ~ q is always, sometimes, never false.

If p is true and q is true, then ~ p -> ~ q is always, sometimes, never true.

If p -> q is true and q is true, then p always, sometimes, never is

Answers

Answer 1
Answer:

1. If p is true and q is false, the p -> q is never true.

2. When p is false and q is true, then p or q is always true.

3. If p is true and ~ q is false, then p -> ~ q is never false.

4. If p is true and q is true, then ~ p -> ~ q is always true.

5. If p -> q is true and q is true, then p is always true.

Further Explanation:

The logic gates are used here.

Here, the symbol -> is for implication. Implication p-> q means that if p is true then q must be true.

So let us look at all the questions one by one.

1. If p is true and q is false, the p -> q is always, sometimes, never true.

p -> q

true -> false

The true should imply true so the given statement will never be true.

2. When p is false and q is true, then p or q is always, sometimes, never true.

false or true

We know that in or gate even if one input is true, the whole output is true. So this statement will be always true given p is false and q is true.

3. If p is true and ~ q is false, then p -> ~ q is always, sometimes, never false.

This translates to:

true -> true

So it will never be false.

4. If p is true and q is true, then ~ p -> ~ q is always, sometimes, never true.

This translates to:

false -> false

This will always be true.

5. If p -> q is true and q is true, then p is always, sometimes, never true.

If p->q is true and q is true then p will always be true. "Implies to" states that in p->q, in order for q to be true p has to be true. So p will always be true.

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Answer 2
Answer:

Answer:

Your answer is: Always true

Step-by-step explanation:

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Angle D is a circumscribed angle of circle O.Circle O is shown. Triangle A C B is inscribed within circle O. Side A B goes through point O. The length of A C is 15 and the length of C B is 8. Angle A C B is a right angle. Line segment O E is a radius. Tangents B D and E D intersect at point D outside of the circle to form kite O B D E. Angles O B D and D E O are right angles. The length of E D is 5.

What is the perimeter of kite OBDE?

17 units
23 units
27 units
60 units

Answers

Since angle D is a circumscribed angle of circle O, the perimeter of kite OBDE is equal to 27 units in accordance with Pythagorean's Theorem.

How to calculate the perimeter of kite OBDE.

In order to determine the perimeter of kite OBDE, we would apply Pythagorean's Theorem because angles OBD and DEO are right angles.

Mathematically, Pythagorean Theorem is given by this formula:

c² = a² + b²

Where:

  • h is the hypotenuse.
  • a is the adjacent side.
  • b is the opposite side.

Given the following data:

Length of AC = 15 units.

Length of CB = 8 units.

Length of ED = 5 units.

Substituting the given parameters into the formula, we have;

c² = 15² + 8²

c² = 225+64

c =√289

c = 17 units.

Since a kite is a quadrilateral with four (4) equal sides and two (2) congruent sides, we can deduce the following:

Side OB = OE = 1/2 × 17 = 8.5 units.

Side BD = ED = 5 units.

Therefore, the perimeter of kite OBDE is given by:

Perimeter = 17 + (5 + 5)

Perimeter = 17 + 10

Perimeter = 27 units.

Read more on Pythagorean Theorem here: brainly.com/question/16176867

Answer:

27

Step-by-step explanation:

Write a fraction that is a multiple of 4/5.

Answers

so if we assume that the ending fraction has to be equal to the origonal fraction then we multiply it by 1 or x/x where x=x

so 4/5 times 2/2=8/10
4/5 times 3/3=12/15
and so on

Can someone explain lesson 6.3 math boxes in everyday math regarding the missing numbers on the number lines

Answers

just subtract the number infront of missing number to the opposing number

Answer:points

Step-by-step explanation:

Difference of numbers. The larger of two integers is 8 less than twice the smaller. When the smaller number is subtracted from the larger, the difference is 17. Find the two numbers.

Answers

let's call the numbers x (the bigger) and y (the smaller)

The larger of two integers is 8 less than twice the smaller. this means:

x=2z-8
When the smaller number is subtracted from the larger, the differenceis 17.

This means: x-z=17.


i'll insert
2z-8

instead of x, since it's equal to it: 
2z-8-z=17

2z-z is just z:
z-8=17

so

z=25.
and x must be 25+17=42

let's check:The larger of two integers is 8 less than twice the smaller.
twice the smaller is 50, and yes, 42 is 8 smaller than 50, it's correct!

so the two numbers are 25 and 42

If you can’t see it’s (2/5)^2

Answers

Answer: 0.16

Step-by-step explanation:

2/5=.4

.4^2=0.16

Answer:

0.16

Step-by-step explanation divide 2 with 5 you get 0.4. After that you have an exponent. You will take 0.4 to the 2nd power

In a class, there are io boys and 15 girls. three students are selected at random. The probability that the selected students are 1 boy and 2 girls, is options: a. 25/36 b. 18/23 c. 21/46 d. 1/32

Answers

Answer:

Step-by-step explanation:

Correct option is C)

There are 15 boys and 10 girls in a class

We have to select 3 students such that there should be 1 girl and 2 boys

The number of ways we can select 3 students is  

25C3=2300

The number of ways we can select 3 students such that there is 1 girl and 2 boys  is 15×7×10=1050

The probability is 1050/2300 =21/46

​

Therefore the correct option is C

Final answer:

finding the number of combinations for the desired scenario and the total possible combinations, we find that the probability is 21/46.

Explanation:

In order to solve this problem, we need to apply the principles of combinatorics and probability. The total number of students in the class is 25 (10 boys and 15 girls). Firstly, let's calculate the combinations for the scenario of selecting 1 boy out of 10. This can be done by 10C1 resulting in 10 possibilities. Secondly, let's calculate the combinations of selecting 2 girls out of 15, which is 15C2 and gives us 105 possibilities.

Multiply those together to find the total scenario we're interested in, which is 1,050. The total possible combinations of selecting 3 students out of 25 irrelevant of gender would be 25C3, resulting in 2,300 possible combinations.

Therefore, the probability that the selected students are 1 boy and 2 girls is 1,050/2,300. Simplifying this fraction gives us 21/46.

Learn more about Combinatorics and Probability here:

brainly.com/question/34324838

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