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Zarrin [17]
3 years ago
13

⦁ Evaluate ⦁ 6! ⦁ 8P5 ⦁ 12C4

Mathematics
2 answers:
lisabon 2012 [21]3 years ago
7 0
6! = 6 * 5 *4*3*2*1 =720

8p5 = 8*7*6*5*4 = 6720

12C4 = (12p4)/4! = (12*11*10*9)/4*3*2*1 = 495
Vesnalui [34]3 years ago
6 0
N! = n(n-1)(n-2)(n-3)....
∴6! = 6×5×4×3×2×1 = 720
⁸P₅ = \frac{8!}{(8-5)!} =  \frac{8*7*6*5*4*3*2*1}{3*2*1} = 6720
¹²C₄ = \frac{12!}{4!*(12-4)!} =  \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4!*8!}
       = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1*8*7*6*5*4*3*2*1} = 495

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Consider the radical equation (3√6-x)+4=-8. Explain why the calculation in Problem 1 does not produce a solution to the equation
murzikaleks [220]

Answer:

See explanation below

Step-by-step explanation:

<u>First we will solve the radical equation</u> (which I guess was problem 1),

Let's start by simplifying it:

3\sqrt{6-x}+4=-8\\ 3\sqrt{6-x}=-8-4\\ 3\sqrt{6-x}=-12\\\sqrt{6-x}=-4

Now we will solve the equation by squaring both sides of the equation:

\sqrt{6-x} =-4\\6-x=-4^{2} \\6-x=16\\-x=16-6\\-x=10\\x=-10

So the calculation for x was that x = -10

However, this does not produce a solution to the equation: When we plug this value into the radical equation we get:

3\sqrt{6-x} +4= -8\\3\sqrt{6-(-10)} +4=-8\\3\sqrt{16}+4 = -8\\ 3(4)+4 = -8\\12 + 4 = -8

This happens because <u>when we first squared both sides of the equation in the first part of the problem we missed one value for x </u>(remember that all roots have 2 answers, a positive one and a negative one) while squares are always positive.

When we squared the root, we missed one value for x and that is why the calculation does not produce a solution to the equation.

6 0
2 years ago
Please help me with this question ☝️☝️
SVETLANKA909090 [29]

Answer:

\frac{-4}{5}

Step-by-step explanation:

\frac{-4}{15}* 3 = -4/5

3 0
2 years ago
Plspslososlslssllsjidpls help
hodyreva [135]

Answer:

Step-by-step explanation:

it's the second answer ,  the number of degrees between 0 and -5

7 0
3 years ago
PLEASE PEOPLE, HELP ME!! Geometry
Oksanka [162]
I use the sin rule to find the area

A=(1/2)a*b*sin(∡ab)

1) A=(1/2)*(AB)*(BC)*sin(∡B)
sin(∡B)=[2*A]/[(AB)*(BC)]

we know that
A=5√3
BC=4
AB=5
then

sin(∡B)=[2*5√3]/[(5)*(4)]=10√3/20=√3/2
(∡B)=arc sin (√3/2)= 60°

 now i use the the Law of Cosines 

c2 = a2 + b2 − 2ab cos(C)

AC²=AB²+BC²-2AB*BC*cos (∡B)

AC²=5²+4²-2*(5)*(4)*cos (60)----------- > 25+16-40*(1/2)=21

AC=√21= 4.58 cms

the answer part 1) is 4.58 cms

2) we know that

a/sinA=b/sin B=c/sinC

and

∡K=α

∡M=β

ME=b

then

b/sin(α)=KE/sin(β)=KM/sin(180-(α+β))

KE=b*sin(β)/sin(α)

A=(1/2)*(ME)*(KE)*sin(180-(α+β))

sin(180-(α+β))=sin(α+β)

A=(1/2)*(b)*(b*sin(β)/sin(α))*sin(α+β)=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

KE/sin(β)=KM/sin(180-(α+β))

KM=(KE/sin(β))*sin(180-(α+β))--------- > KM=(KE/sin(β))*sin(α+β)

the answers part 2) are

side KE=b*sin(β)/sin(α)
side KM=(KE/sin(β))*sin(α+β)
Area A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

5 0
2 years ago
Which of the following is most likely the next step in the series
Anon25 [30]
<h3>Answer: Choice D</h3>

=============================

Explanation:

The first figure has 4 sides (quadrilateral)

The second figure has 5 sides (pentagon)

The third figure has 6 sides (hexagon)

Each time we increase the number of sides by 1.

The fourth figure must have 7 sides (heptagon). The only thing that has 7 sides is the figure listed in choice D.

6 0
3 years ago
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