PLEASE ANSWER THIS QUESTION

Vera_Pavlovna [14]

Step-by-step explanation:

here's your solution

=> in first figure , base = 5.5, perpendicular =7.8

=> h^2 = 5.5^2 + 7.8^2

=> h^2 = 30.25 + 60.84

=>h^2 = 91.09

=> h = √91.09

=> h = 9.5

 Both figure are congruent

enc we will get a rectangle by add both 

hope it helps

3 0

3 years ago

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Find the area of the triangle

UNO [17]

Answer:

can u show a little clear pic with full triangle shown

6 0

3 years ago

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A card is chosen from a standard deck of cards. What is the probability that the card is a club, given that the card is black?

leonid [27]

From a standard deck of cards, one card is drawn. What is the probability that the card is black and a jack? P(Black and Jack) P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26 A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen or an ace. P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13 WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the probability that they will both be aces? P(AA) = (4/52)(3/51) = 1/221. 1 WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a king? P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed. WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the probability of drawing the first queen which is 4/52. The probability of drawing the second queen is also 4/52 and the third is 4/52. We multiply these three individual probabilities together to get P(QQQ) = P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible. Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit) What's the probability of being dealt a royal flush in a five card hand from a standard deck of cards? (Note: A royal flush is a 10, Jack, Queen, King, and Ace of the same suit. A standard deck has 4 suits, each with 13 distinct cards, including these five above.) (NB: The order in which the cards are dealt is unimportant, and you keep each card as it is dealt -- it's not returned to the deck.) The probability of drawing any card which could fit into some royal flush is 5/13. Once that card is taken from the pack, there are 4 possible cards which are useful for making a royal flush with that first card, and there are 51 cards left in the pack. therefore the probability of drawing a useful second card (given that the first one was useful) is 4/51. By similar logic you can calculate the probabilities of drawing useful cards for the other three. The probability of the royal flush is therefore the product of these numbers, or 5/13 * 4/51 * 3/50 * 2/49 * 1/48 = .00000154

5 0

3 years ago

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What is the standard form of y=2x+3

saw5 [17]

$2 \times x + 3 = y \\$
I believe but I am not too sure.

8 0

3 years ago

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The height of a triangle is 4 sqrt 3 . What is the perimeter of the equatorial triangle?

Karo-lina-s [1.5K]

The perimeter of the equatorial triangle is 24 units

<h3>Solution:</h3>

Given that,

An equilateral triangle has an height equal to $4 \sqrt{3}$

The triangle is shown below

From Triangle ABC in the shown figure AD $=4 \sqrt{3}$

Let the sides of the equilateral triangle be ‘a’

AB = BC = a

Since, it is an equilateral triangle we get,

BD = DC = a ÷ 2

Now, using Pythagoras Theorem in Triangle ABD,

The Pythagorean theorem is this: In a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse.

$\mathrm{AB}^{2}=\mathrm{BD}^{2}+\mathrm{AD}^{2}$