Given:
Area of a square frame = 81 square inches.
To find:
Perimeter of the frame.
Solution:
Let length of each side of the frame = a inches.
Area of square frame is
Taking square root on both sides,
Side length cannot be negative. So, a=9 inches.
Perimeter of frame is
Therefore, the perimeter is 36 inches.
The perimeter of the frame is 36 inches.
To find the perimeter of a square, we need to know the length of its sides. Since the area of the frame is given as 81 square inches, we can find the length of one side by taking the square root of the area. The square root of 81 is 9, so each side of the frame measures 9 inches.
Since a square has four equal sides, we can find the perimeter by multiplying the length of one side by 4. Therefore, the perimeter of the frame is 9 inches x 4 = 36 inches.
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Answer:
The positive solution is x=9
Step-by-step explanation:
Given the equation
we have to find the positive solution of above equation.
By splitting middle-term method
The solution is x=-4 and x=9
Hence, the positive solution is x=9
Answer:
Correct answer is option D) drawing the 6 of clubs or the 6 of spades
Step-by-step explanation:
Given that:
A standard deck of 52 playing cards with 4 suits.
A be the event of drawing a card with 6 from the deck.
B be the event of drawing a black card from the deck.
So, event A will have 4 possibilities i.e. {6 of club, 6 of spade, 6 of diamond, 6 of heart}
And event B will have 26 possibilities {Any card (including 6) from club or spade}
Intersection of two sets is defined as the common elements in the two sets.
As per the explanation of the sets and elements in the sets given above:
If we take intersection it will be:
{6 of club or 6 of spade}
Hence, Correct answer is option D) drawing the 6 of clubs or the 6 of spades
The intersection of events A and B is represented by drawing the 6 of clubs or the 6 of spades (D) .
The intersection of events A (drawing a 6) and B (drawing a black playing card) refers to the cards that satisfy both criteria. In this case, event A consists of the card 6 from any suit, while event B consists of the black cards (spades and clubs) from any value. To determine the intersection, we need to find the cards that are both a 6 and black.
Looking at the options given, option D) drawing the 6 of clubs or the 6 of spades represents the cards that satisfy both events. The 6 of clubs and the 6 of spades are black cards and also have a value of 6. Therefore, the intersection of events A and B is represented by option D).
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books.
6.07
c.
6.7
b.
6.17
d.
6.71
Please select the best answe