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2 years ago

I need to find the answer to this question and please explain how.

Mathematics

1 answer:

deff fn [24]2 years ago

8 0

Among the given statements, the true statements are:

If a number is divisible by 3 and 4, then it is divisible by 12.
The number 119 is a composite number.

In the question, we are asked for the true statements among the given options:

1. All even numbers are composite numbers.

Answer - False

Reason - 2 is an even number, which is not a composite number.

2. If a number is divisible by 3 and 4, then it is divisible by 12.

Answer - True

Reason - If any number, n, is divisible by any two numbers, x and y, then it is also divisible by their LCM, and the LCM of 3 and 4 is 12.

3. All prime numbers are odd numbers.

Answer - False

Reason - 2 is a prime number, which is not odd.

4. The number 119 is a composite number.

Answer - True

Reason - The number 119 is a composite number, as it is not a prime number, as it is divisible by 7 and 17 too, other than 1 and itself.

5. 12 is a factor of 6.

Answer - False.

Reason - When 6 is divided by 12, the integral quotient that we get is 0, and the remainder is 6, but for any number x, to be a factor of y, on dividing y by x, the remainder should be 0.

6. If a number is divisible by 6 and 4, then it is also divisible by 4.

Answer - False

Reason - If any number, n, is divisible by any two numbers, x and y, then it is also divisible by their LCM, and the LCM of 6 and 4 is 12.

Thus, a number divisible by 6 and 4 is divisible by 12 but not necessarily by 24. Example 12 is divisible by both 6 and 4 but not divisible by 24.

Thus, among the given statements, the true statements are:

If a number is divisible by 3 and 4, then it is divisible by 12.
The number 119 is a composite number.

Learn more about divisibility at

brainly.com/question/26789053

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Step-by-step explanation:

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3 years ago

Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt

Leni [432]

Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = $\sqrt{Δx^2+Δy^2}$
d₁ = $\sqrt{(-2-1)^2+(1-4)^2}$
d₂ = $\sqrt{(x-1)^2+(y-4)^2}$
d₁ = d₂
∴ $\sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2}$ ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = $\pm \sqrt{9} = \pm 3$
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1) or (-2,7)

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