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tensa zangetsu [6.8K]
3 years ago
8

Hey! i need help plz

Mathematics
1 answer:
tia_tia [17]3 years ago
6 0
X=6 and y=-9. Not sure if y is correct since I kind of rushed but I hope this helps!
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What’s the correct answer for this?
DiKsa [7]
48 degrees

The triangles are congruent, so if angle B is between side lengths 6 and 6.2, it’s going to be the same for the other triangle
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10 POINTS!!: ANSWER THIS QUESTION<br> (please no bad answers)
nevsk [136]

Answer:

Write in standard form.

3x+2y=8

Step-by-step explanation:

8 0
3 years ago
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Examine the quadratic equation 9x^2+12x +4= 0.
mestny [16]

Answer:

See answers and explanations

Step-by-step explanation:

A) Discriminant = b² - 4ac = (12)² - 4(9)(4) = 144 - 144 = 0

B) Since the discriminant is 0, this means the equation only has one root.

C) The root is real in this case since the discriminant is not negative. Otherwise, we would have two imaginary roots if it were (remember that complex roots come in pairs)

4 0
3 years ago
The perimeter of a​ standard-sized rectangular rug is 4040 ft. the length is 22 ft longer than the width. find the dimensions.
Nataly_w [17]
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2(l+w) = 4040
2(w+22+w) = 4040
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3 0
3 years ago
Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select
Blizzard [7]

Answer:

Probability is:   $ \frac{\textbf{13}}{\textbf{51}} $

Step-by-step explanation:

From a deck of 52 cards there are 26 black cards. (Spades and Clubs).

Also, there are 26 red cards. (Hearts and Diamonds).

First, we determine the probability of drawing a black card.

P(drawing a black card) = $ \frac{number \hspace{1mm} of  \hspace{1mm} black  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm} of  \hspace{1mm} cards} $  $ = \frac{26}{52} = \frac{\textbf{1}}{\textbf{2}} $

Now, since we don't replace the drawn card, there are only 51 cards.

But the number of red cards is still 26,

∴ P(drawing a red card) = $ \frac{number  \hspace{1mm} of  \hspace{1mm} red  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm}of  \hspace{1mm} cards} $  $ = \frac{26}{51}  $

Now, the probability of both black and red card = $ \frac{1}{2} \times \frac{26}{51} $

$ = \frac{\textbf{13}}{\textbf{51}} $

Hence, the answer.

5 0
3 years ago
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