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3 years ago

15

There are 12 face cards in a standard deck of 52 cards. How many ways can you arrange a standard deck of 52 cards such that the

first card is a face card?

1 answer:

evablogger [386]3 years ago

8 0

Answer:

Total number of required ways = 12 × 51!

Step-by-step explanation:

Number of total cards in the deck = 52

Number of face cards in the deck = 12

Now, we need to find the number of ways such that the first card is always a face card

Now, first card is face card so number of cards left to be arranged = 52 - 1 = 51

Number of ways to arrange 51 cards = 51!

Also, Number of face cards = 12

So, Total number of required ways = 12 × 51!

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At a zoo, an Asian elephant is about 3 tons and eats about 300 pounds of food a day what percent of its body weight does the ele

olchik [2.2K]

3 Tons = 6000 pounds
300 is 5% of 6000 so the answer is 5% :)

8 0

3 years ago

Y=-x^2+6x-4<br> Find the Axis of Symmetry and the Vertex, Also solve the whole equation.

Inessa [10]

Answer:

1. <em>Axis of symmetry</em>: x = 3

2. <em>Vertex:</em> (3,5)

3. <em>Solution of the equation</em>:

<u />

x-intercepts:

$(3-\sqrt{5},0) \\\\(3+\sqrt{5},0)$

y-intercept: (0, -4)

Explanation:

<u>1. Equation:</u>

$y=-x^2+6x-4$

<u>2. </u><em><u>Axis of symmetry:</u></em>

That is the equation of a parabola, whose standard form is:

$y=ax^2+bx+c$

Where:

$a=-1;b=6;c=-4$

The axis of symmetry is the vertical line with equation:

$x=-b/2a$

Substitute $a=-1,\text{ and }b=6$

$x=-6/[(2)(-1)]=-6/(-2)=3$

Thus, the axis of symmetry is:

$x=3$

<em><u>3. Vertex</u></em>

<em><u /></em>

The x-coordinate of the vertex is equal to the axys of symmetry, i.e x = 3.

To find the y-xoordinate, substitute this value of x into the equation for y:

$y=-x^2+6x-4\\\\y=-(3)^2+6(3)-4=-9+18-4=5$

Therefore, the vertex is (3, 5)

<u>4. Find the x-intercepts</u>

The x-intercepts are the roots of the equation, which are the points wher y = 0.

$y=-x^2+6x-4=0$

Use the quadratic equation:

$x=\frac{-b\pm \sqrt{b^2-4ac} }{2a} \\\\x=\frac{-6\pm\sqrt{(-6)^2-4(-1)(-4)} }{2(-1)}\\\\x_1=3-\sqrt{5} \\\\x_2=3+\sqrt{5}$

<u>5. Find the y-intercept</u>

<u />

The y-intercet is the value of y when x=0:

$y=-x^2+6x-4\\\\y=-(0)^2+6(0)-4=-4$

7 0

4 years ago

Expand the given power by using Pascal’s triangle. (9a - 10b)^6

xxMikexx [17]

Answer:

$531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

Step-by-step explanation:

1 n=0

1 1 n=1

1 2 1 n=2

1 3 3 1 n=3

1 4 6 4 1 n=4

1 5 10 10 5 1 n=5

1 6 15 20 15 6 1 n=6

This is where n is the exponent in

$(x+y)^n$ .

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

Now we want to expand:

$(9a-10b)^6$ or we we can rewrite as $(9a+(-10b))^6$ .

Let's replace $x$ with $(9a)$ and $y$ with $(-10b)$ in the expansion:

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

$((9a)+(-10b))^6$

$=1(9a)^6+6(9a)^5(-10b)+15(9a)^4(-10b)^2+20(9a)^3(-10b)^3+15(9a)^2(-10b)^4+6(9a)(-10b)^5+1(-10b)^6$

Let's simplify a bit:

$=9^6a^6-60(9)^5a^5b+15(-10)^2(9)^4a^4b^2+20(9)^3(-10)^3a^3b^3+15(9)^2(-10)^4a^2b^4+6(9)(-10)^5ab^5+(-10)^6b^6$

$=531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

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4 years ago

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umka2103 [35]

Answer:

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Step-by-step explanation:

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3 years ago

Need help ASAP ‼️<br> The subject is trigonometric ratios.

postnew [5]

Answer:

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3 years ago