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3 years ago

20 Points!!

Mathematics

2 answers:

shepuryov [24]3 years ago

7 0

Answer:

D) { x | x < 5}

Step-by-step explanation:

2(4 x + 3) < 5 x + 21

Distribute

8x +6 < 5x+21

Subtract 5x from each side

8x-5x+6 < 5x-5x+21

3x+6 < 21

Subtract 6 from each side

3x+6-6<21-6

3x <15

Divide each side by 3

3x/3 <15/3

x < 5

madam [21]3 years ago

4 0

Answer:

Step-by-step explanation:

8x-5x<21-6

x<15÷3

x<5

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If m AC =86°, find m <AVC

bulgar [2K]

M<AVC = 43 degrees, because it is the angle's measure is 1/2 of the arc's

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3 years ago

Write a real world situation to explain -$15<$7

Ronch [10]

Jeff has -15$ his gf faith gives him 7.

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3 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

3 years ago

Read 2 more answers

Expanded form for 48,243. (

Dima020 [189]

(4*10,000)+(8*1,000)+(2*100)+(4*10)+(3*1)

40,000+8,000+200+40+3=48,243

That is one way. You can also do this using exponential form!

Hope that helps

4 0

3 years ago

How cos π/3=1/2?<br>having some problems learning continuity.

Bingel [31]

You can show that $\cos\frac\pi3=\frac12$ by constructing a triangle.

Take two points, O(0, 0) and A(1, 0), and let B be the point on the unit circle such that the angle between the line segments OA and OB is $\frac\pi3$ radians.

Since both A and B lie on the circle, the line segments OA and OB both have length 1 (same as the circle's radius). We finish constructing the triangle by connect A and B.

Since OB and OA have the same length, triangle OAB is isosceles, but more than that, it's also equilateral. Why? Because the interior angles of any triangle always add to $\pi$ radians. We know one of the angles is $\frac\pi3$ radians, which leaves a contribution of $\frac{2\pi}3$ radians between the remaining angles A and B. Angles A and B must be congruent (because OAB is isosceles), which means they also have measure $\frac\pi3$ radians.

Next, draw an altitude of the triangle through point B, and label the point where it meets the "base" OA, C. Since OAB is equilateral, the altitude BC is also a perpendicular bisector. That means OC has length $\frac12$ , and by definition of $\cos$ we have

$\cos\dfrac\pi3=\dfrac{OC}{OB}=\dfrac{\frac12}1=\dfrac12$

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3 years ago