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Mrac [35]
2 years ago
6

There are 20 machines at the gym and Manuel wants to perform exercises on 5 different machines while he is there. How many diffe

rent ways can he select the machines to use?
31008

1860480

100

15504
Mathematics
2 answers:
sasho [114]2 years ago
7 0
  • Total machines: 20
  • Required: on 5 <u>different</u> machines

Selection of machines is persons choice and not arranged specifically, thus we will use Combination.

⇒ 20C5

⇒ 20!/5!(20-5)!

⇒ 15504

Brut [27]2 years ago
3 0
  • n=20
  • r=5

Use combination

total ways

\\ \rm\dashrightarrow ^nC_r=\dfrac{n!}{r!(n-r)!}

\\ \rm\dashrightarrow ^{20}C_5

\\ \rm\dashrightarrow \dfrac{20!}{5!15!}

\\ \rm\dashrightarrow \dfrac{20(19)(18)(17)(16)(15!)}{5(4)(3)(2)(15!}

\\ \rm\dashrightarrow 15504

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Drag each tile to the correct box. Triangle ABC is inscribed in a circle centered at point o. CK630 B 1180 Order the measures of
Rudik [331]

Answer:

AC < BC < AB

Step-by-step explanation:

Given: BC = 118

AB = 63*2 = 126

AC = 360 - (BC+AB)

AC = 360 - (118+126) = 116

6 0
2 years ago
Find the vertices and foci of the hyperbola. 9x2 − y2 − 36x − 4y + 23 = 0
Xelga [282]
Hey there, hope I can help!

NOTE: Look at the image/images for useful tips
\left(h+c,\:k\right),\:\left(h-c,\:k\right)

\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:a\:right-left\:facing:H}
with the center of (h, k), semi-axis a and semi-conjugate - axis b.
NOTE: H = hyperbola

9x^2-y^2-36x-4y+23=0 \ \textgreater \  \mathrm{Subtract\:}23\mathrm{\:from\:both\:sides}
9x^2-36x-4y-y^2=-23

\mathrm{Factor\:out\:coefficient\:of\:square\:terms}
9\left(x^2-4x\right)-\left(y^2+4y\right)=-23

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1
\frac{1}{1}\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x^2-4x+4\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y+4\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

\mathrm{Refine\:}-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right) \ \textgreater \  \frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=1 \ \textgreater \  Refine
\frac{\left(x-2\right)^2}{1}-\frac{\left(y+2\right)^2}{9}=1

Now rewrite in hyperbola standardform
\frac{\left(x-2\right)^2}{1^2}-\frac{\left(y-\left(-2\right)\right)^2}{3^2}=1

\mathrm{Therefore\:Hyperbola\:properties\:are:}\left(h,\:k\right)=\left(2,\:-2\right),\:a=1,\:b=3
\left(2+c,\:-2\right),\:\left(2-c,\:-2\right)

Now we must compute c
\sqrt{1^2+3^2} \ \textgreater \  \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \  1^2 = 1 \ \textgreater \  \sqrt{1+3^2}

3^2 = 9 \ \textgreater \  \sqrt{1+9} \ \textgreater \  \sqrt{10}

Therefore the hyperbola foci is at \left(2+\sqrt{10},\:-2\right),\:\left(2-\sqrt{10},\:-2\right)

For the vertices we have \left(2+1,\:-2\right),\:\left(2-1,\:-2\right)

Simply refine it
\left(3,\:-2\right),\:\left(1,\:-2\right)
Therefore the listed coordinates above are our vertices

Hope this helps!

8 0
3 years ago
!!!!!!!!!! please help
Sergio [31]

Answer:

310 pages in book = 100%

Step-by-step explanation:

5 0
3 years ago
Rework problem 35 from the Chapter 2 review exercises in your text, involving auditioning for a play. For this problem, assume 9
sweet-ann [11.9K]

Answer: 1) 6300 ways

2) 2520 ways

3) 0.067

Step-by-step explanation:

For this problem, assume 9 males audition, one of them being Winston, 5 females audition, one of them being Julia, and 6 children audition. The casting director has 3 male roles available, 1 female role available, and 2 child roles available.

How many different ways can these roles be filled from these auditioners?

Available: 9M and  5F and 6C

Cast: 3M and 1F and  2C

As it is not ordered: C₉,₃ * C₅,₁ * C₆,₂

C₉,₃ = 9!/3!.6! = 84

C₅,₁ = 5!/1!.4! = 5

C₆,₂ = 6!/2!.4! = 15

C₉,₃ * C₅,₁ * C₆,₂ = 84.5.15 = 6300

How many different ways can these roles be filled if exactly one of Winston and Julia gets a part?

2 options Winston gets or Julia gets it:

1) Winston gets it but Julia no:

8 male for 2 spots

4 females for 1 spot

6 children for 2 spots

C₈,₂ * C₄,₁ * C₆,₂

C₈,₂ = = 8!/2!.6! = 28

C₄,₁ = 4!/1!.3! = 4

C₆,₂ = 6!/2!.4! = 15

C₈,₂ * C₄,₁ * C₆,₂ = 28.4.15 = 1680

2) Julia gets it but Winston does not

8 male for 3 spots

1 female for 1 spot

6 children for 2 spots

C₈,₃ * C₆,₂

C₈,₃ = 8!/3!.5! = 56

C₆,₂ = 6!/2!.4! = 15

C₉,₃ * C₆,₂ = 56.15 = 840

1) or 2) = 1) + 2) = 1680 + 840 = 2520

What is the probability (if the roles are filled at random) of both Winston and Julia getting a part?

8 male for 2 spots

1 female for 1 spot

6 children for 2 spots

C₈,₂ * C₆,₂

C₈,₂ = = 8!/2!.6! = 28

C₆,₂ = 6!/2!.4! = 15

C₈,₂ * C₆,₂ = 28.15 = 420

p = 420/6300 = 0.067

6 0
3 years ago
Y=x^2-12x+45 vertex form and coordinate vertex
zysi [14]
Best Answer

<span><span> x2-12x-45=0</span> </span>Two solutions were found :<span> x = 15 x = -3</span>

Step by step solution :<span>Step  1  :</span>Skip Ad
Trying to factor by splitting the middle term

<span> 1.1 </span>    Factoring <span> x2-12x-45</span> 

The first term is, <span> <span>x2</span> </span> its coefficient is <span> 1 </span>.
The middle term is, <span> -12x </span> its coefficient is <span> -12 </span>.
The last term, "the constant", is <span> -45 </span>

Step-1 : Multiply the coefficient of the first term by the constant <span> <span> 1</span> • -45 = -45</span> 

Step-2 : Find two factors of  -45  whose sum equals the coefficient of the middle term, which is  <span> -12 </span>.

<span><span>     -45   +   1   =   -44</span><span>     -15   +   3   =   -12   That's it</span></span>


Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -15  and  3 
                     <span>x2 - 15x</span> + 3x - 45

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x • (x-15)
              Add up the last 2 terms, pulling out common factors :
                    3 • (x-15)
Step-5 : Add up the four terms of step 4 :
                    (x+3)  •  (x-15)
             Which is the desired factorization

<span>Equation at the end of step  1  :</span> (x + 3) • (x - 15) = 0 <span>Step  2  :</span>Theory - Roots of a product :

<span> 2.1 </span>   A product of several terms equals zero.<span> 

 </span>When a product of two or more terms equals zero, then at least one of the terms must be zero.<span> 

 </span>We shall now solve each term = 0 separately<span> 

 </span>In other words, we are going to solve as many equations as there are terms in the product<span> 

 </span>Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

<span> 2.2 </span>     Solve  :    x+3 = 0<span> 

 </span>Subtract  3  from both sides of the equation :<span> 
 </span>                     x = -3 

Solving a Single Variable Equation :

<span> 2.3 </span>     Solve  :    x-15 = 0<span> 

 </span>Add  15  to both sides of the equation :<span> 
 </span>                     x = 15 

Supplement : Solving Quadratic Equation Directly<span>Solving <span> x2-12x-45</span>  = 0 directly </span>

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

<span> 3.1 </span>     Find the Vertex of   <span>y = x2-12x-45

</span>Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero).<span> 

 </span>Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.<span> 

 </span>Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.<span> 

 </span>For any parabola,<span>Ax2+Bx+C,</span>the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is   6.0000 <span> 

 </span>Plugging into the parabola formula   6.0000  for  x  we can calculate the  y -coordinate :<span> 
 </span><span> y = 1.0 * 6.00 * 6.00 - 12.0 * 6.00 - 45.0 
</span>or   y = -81.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : <span> y = x2-12x-45</span>
Axis of Symmetry (dashed)  {x}={ 6.00} 
Vertex at  {x,y} = { 6.00,-81.00}  
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-3.00, 0.00} 
Root 2 at<span>  {x,y} = {15.00, 0.00}</span>

3 0
3 years ago
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