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DedPeter [7]
3 years ago
9

A deck of cards contains 5 black cards numbered 5 to 9 and 5 red cards numbered 1 to 5. How many 6-card hands having 4 black car

ds and 2 red cards can be dealt
Mathematics
1 answer:
tester [92]3 years ago
5 0

Answer:

50

Step-by-step explanation:

4 black cards = 5C4 = 5

2 red cards = 5C2 = 10

5 x 10 = 50

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The three trigonometric functions can be used to calculate unknown side lengths and angles in which type of triangle(s)?
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Factor perfect square trinomials
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Divide 2y from all terms
2y(y^2+6y+9)
4 0
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Forty-five people volunteered, but 200% of that number are needed. How<br> many people are needed?
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8 0
3 years ago
Note that WXYZ has vertices W(-1, 2), X(-5, 7), y(-1, -2), and Z (3, -7).
svp [43]

a. Slope of WZ = -2.25; Slope of WX = 5

b. WZ = √97; WX = √41

c. WXYZ is not a <em>rectangle, rhombus, nor a square</em>. We can conclude that: <em>D. WXYZ is none of these</em>.

<h3>Slope of a Segment</h3>

Slope = change in y/change in x

Given:

W(-1, 2), X(-5, 7), Y(-1, -2), and Z (3, -7)

a. Slope of WZ and slope of WX:

Slope of WZ = (-7 - 2)/(3 -(-1)) = -2.25

Slope of WX = (7 - 2)/(-1 -(-1)) = 5

b. Use distance formula, d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, to find WZ and WX:

WZ = \sqrt{(3 - (-1))^2 + (-7 - 2)^2}\\\\\mathbf{WZ = \sqrt{97} }

WX = \sqrt{(-5 -(-1))^2 + (7 - 2)^2}\\\\\mathbf{WX = \sqrt{41} }

c. The quadrilateral WXYZ have adjacent sides that are not perpendicular to each other and have different slopes and different lengths, so therefore, WXYZ is not a rectangle, rhombus, nor a square. We can conclude that: <em>D. WXYZ is none of these</em>.

Learn more about slopes on:

brainly.com/question/3493733

5 0
3 years ago
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