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Irina-Kira [14]
3 years ago
10

If five kids sit on a d and one has an orgasm how many are still going

Mathematics
1 answer:
Ierofanga [76]3 years ago
4 0

Answer:

jesus...you're twisted anyway 69 :)

Step-by-step explanation:

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Determine whether the following pairs of lines are parallel,perpendicular, or neither. Y=3x-2 -3x+y=5
Bad White [126]
-3x+y=5\\
y=3x+5

The slopes of both lines are equal so they're parallel.
5 0
4 years ago
A zoo keeper wants to fence a rectangular habitat for goats. The length of the habitat should be at least 80 feet and the perime
zimovet [89]

Answer:

Step-by-step explanation:

Let x ft be the length of the fence and y ft be the width of the fence.

1. The length of the habitat should be at least 80 feet, then

2. The perimeter of the habitat should be no more than 300 feet. The perimeter of the fences 2x+2y ft, then

3. The system of inequalities that represent the model is

The graph of these inequalities is attached.

Two possible solutions are

1. x=100 ft, y=20 ft (100≥80 and 2·100+2·20=240≤300);

2. x=110 ft, y=30 ft (110≥80 and 2·110+2·30=280≤300).

7 0
3 years ago
Which of the following is the multiplicative inverse of 35/9
labwork [276]
The multiplicative inverse of something is a fancy way to say reciprocal.
A reciprocal is a fancy way to describe the action of flipping the fraction.
So flip 35/9 to get 9/35.
That's the multiplicative inverse.
5 0
3 years ago
TRUE/FALSE. the lengths of the altitude of a hypotenuse is the geometric mean of the lengths of the segments of each of the legs
vitfil [10]

In Right Angled Triangle the lengths of the altitude of a hypotenuse is the geometric mean of the lengths of the segments of each of the legs is TRUE

What is a Right angled triangle?

A triangle is said to be right-angled if one of its angles is exactly 90 degrees. The total of the other two angles is 90 degrees. Perpendicular and the triangle's base are the sides that make up the right angle. The longest of the three sides, the third side is known as the hypotenuse.

The right triangle's three sides are interconnected. The Pythagorean Theorem explains this connection. This theorem states that the Hypotenuse2 = Perpendicular2 + Base2.

  • The right angle, or 90°, is always one angle.
  • The hypotenuse is the side with the 90° angle opposite.
  • The longest side is always the hypotenuse.
  • The other two inner angles add up to 90 degrees.

Let assume a right-angled triangle ABC, right-angled at A.

Let AD be the perpendicular drawn from vertex A on hypotenuse BC, intersecting BC at D.

Now, In right-angle triangle ABC,

By using Pythagoras Theorem, we have

AB^2 +AC^2 = BC^2

Now, In right-angle triangle ABD

Using Pythagoras Theorem, we have

AB^2 =AD^2 + BD^2

Now, In right-angle triangle ACD,

AC^2 = AD^2+CD^2

On adding equation (2) and (3), we get

AB^2 +AC^2 = 2AD^2+ BD^2 +CD^2

Now, using equation (1), the above equation can be rewritten as

2AD^2 +CD^2+BD^2 = BC^2

can be further rewritten as

2AD^2 +BD^2 +CD^2 = (CD+BD)^2\\\\BD^2 +CD^2 +2(BD)(CD)=2AD^2 +BD^2 +CD^2\\\\(BD)(CD)=AD^2

AD is geometric mean of BD and CD

Learn more about Right angled triangle from the link below

brainly.com/question/3770177

#SPJ4

8 0
2 years ago
Given a function I and a subset A of its domain, let I(A) represent the range of lover the set A; that is, I(A) = {I(x) : x E A}
Ede4ka [16]

Step-by-step explanation:

(a)

Using the definition given from the problem

f(A) = \{x^2  \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2  \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4]  = f(A \cap B)\\

Therefore it is true for intersection. Now for union, we have that

A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]

Therefore, for this case, it would be true that f(A\cup B) = f(A)\cup f(B).

(b)

1 is not a set.

(c)

To begin with  

A\cap B \subset A,B

Therefore

g(A\cap B) \subset g(A) \cap g(B)

Now, given an element of g(A) \cap g(B) it will belong to both sets, therefore it also belongs to g(A\cap B), and you would have that

g(A)\cap g(B) \subset  g(A \cap B), therefore  g(A)\cap g(B)  =  g(A \cap B).

(d)

To begin with A,B  \subset A \cup B, therefore

g(A) \cup g(b) \subset g(A\cup B)

7 0
4 years ago
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