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2 years ago

You are going to draw 4 cards from a standard deck of 52 cards. How many different combinations of cards could you draw?

Mathematics

2 answers:

zloy xaker [14]2 years ago

5 0

Answer:

Step-by-step explanation:

To start with we can consider how many possibilities there are for the first card which is obviously 52, as any one of the cards in the deck can be found on the top of the pile. The next card is a bit trickier as there are only 51 possible cards that it could be as the first card takes up a spot. The third card now only has 50 possible cards it could and this continues all the way down to the last card.

is that good

Firlakuza [10]2 years ago

5 0

You can make 208 different combinations

Explanation: To find the combinations you can multiply 52 and 4 and you should get 208

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Suppose cos(x)= -1/3, where π/2 ≤ x ≤ π. What is the value of tan(2x). EDGE

AVprozaik [17]

Answer:

Step-by-step explanation:

We are given that:

$\displaystyle \cos x = -\frac{1}{3}\text{ where } \pi /2 \leq x \leq \pi$

And we want to find the value of tan(2<em>x</em>).

Note that since <em>x</em> is between π/2 and π, it is in QII.

In QII, cosine and tangent are negative and only sine is positive.

We can rewrite our expression as:

$\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}$

Using double angle identities:

$\displaystyle \tan(2x)=\frac{2\sin x\cos x}{\cos^2 x-\sin^2 x}$

Since cosine relates the ratio of the adjacent side to the hypotenuse and we are given that cos(<em>x</em>) = -1/3, this means that our adjacent side is one and our hypotenuse is three (we can ignore the negative). Using this information, find the opposite side:

$\displaystyle o=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$

So, our adjacent side is 1, our opposite side is 2√2, and our hypotenuse is 3.

From the above information, substitute in appropriate values. And since <em>x</em> is in QII, cosine and tangent will be negative while sine will be positive. Hence:

<h2> $\displaystyle \tan(2x)=\frac{2(2\sqrt{2}/3)(-1/3)}{(-1/3)^2-(2\sqrt{2}/3)^2}$ </h2>

Simplify:

$\displaystyle \tan(2x)=\frac{-4\sqrt{2}/9}{(1/9)-(8/9)}$

Evaluate:

$\displaystyle \tan(2x)=\frac{-4\sqrt{2}/9}{-7/9} = \frac{4\sqrt{2}}{7}$

The final answer is positive, so we can eliminate A and B.

We can simplify D to:

$\displaystyle \frac{2\sqrt{8}}{7}=\frac{2(2\sqrt{2}}{7}=\frac{4\sqrt{2}}{7}$

So, our answer is D.

7 0

3 years ago

How many pieces of celery can you fill with peanut butter if you have3\4 cup of peanut an! each piece of celery will hold1\12 cu

Masja [62]

You would be able to fill 9 celery pieces

$\frac{1}{12} * \frac{9}{1} = \frac{9}{12}$ which simplifies to $\frac{3}{4}$

3 0

3 years ago

Which of the statements about the following quadratic equation is true?

Sedaia [141]

After manipulating above equation we get, 2x^2 - 7x - 8= 0. Discriminant= b^2 - 4ac = (-7)^2 - 4(2)(-8) = 113>0. So there are 2 real roots :)

3 0

3 years ago

Read 2 more answers

The perimeter of equilateral triangle ABC is 81/3 centimeters, find the length of the radius and apothem.

MAXImum [283]

There is a typo error, the perimeter of equilateral triangle ABC is 81/√3 centimeters.

Answer:

Radius = OB= 27 cm

Apothem = 13.5 cm

A diagram is attached for reference.

Step-by-step explanation:

Given,

The perimeter of equilateral triangle ABC is 81/√3 centimeters.

Substituting this in the formula of perimeter of equilateral triangle = $3\times\ side$

$3\times\ side$ $=[tex]81\sqrt{3}$

$Side = \frac{81\sqrt{3} }{3} =27\sqrt{3} \ cm$

Thus from the diagram , Side $AB=BC=AC= 27\sqrt{3} \ cm$

We know each angle of an equilateral triangle is 60°.

From the diagram, OB is an angle bisector.

Thus $\angle OBC = 30$ °

Apothem is the line segment from the mid point of any side to the center the equilateral triangle.

Therefore considering ΔOBE, and applying tan function.

$tan\theta =\frac{perpendicular}{base} \\tan\theta=\frac{OE}{BE} \\tan\theta=\frac{OE}{\frac{27\sqrt{3}}{2} } \\tan30\times {\frac{27\sqrt{3} }{2} }= OE\\\frac{1}{\sqrt{3} } \times\frac{27\sqrt{3} }{2} =OE\\$

Thus ,apothem OE= 13.5 cm

Now for radius,

We consider ΔOBE

$cos\theta=\frac{base}{hypotenuse} \\cos30= \frac{BE}{OB} \\Cos30 = \frac{\frac{27\sqrt{3} }{2}}{OB} \\OB= \frac{\frac{27\sqrt{3} }{2}}{cos30} \\OB= \frac{\frac{27\sqrt{3} }{2}}{\frac{\sqrt{3} }{2} } \\OB =27 \ cm$

Thus for

Perimeter of equilateral triangle ABC is 81/√3 centimeters,

The radius of equilateral triangle ABC is 27 cm

The apothem of equilateral triangle ABC is 13.5 cm

4 0

3 years ago

What is the coefficient of 4x^2 ?

aleksley [76]

It has a coefficient of 4 :)

3 0

2 years ago