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kow [346]
3 years ago
5

Suppose a certain religious university has a ratio of approximately 90 males to 125 females enrolled as a students. If he curren

t enrollment at the university is 15000 and every male on campus asks a female student to a dance, how many women will be without dates to the dance?
Mathematics
1 answer:
Nitella [24]3 years ago
7 0
How many males and how many females would there be if the total enrollment is 15000?

Ratio is 

  90 males      15000 - x
  ------------- = ---------------
   125 fem              x

Then 90x = 1875000 - 125x, or 215x = 1875000

                                                      x = approx 8720 females
                       and 15000-8720, or 6279 males

If every male finds a dance partner, that would leave (8720-6279) females, or 2440, without dance partners.  ;(

Check:  is the male:female ratio near 90:125?

Is 6279:8720 approx equal to 90:125?

Yes.  0.72 (approx) = 0.72 (approx).
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what is the equation in point slope form of a line that passes through the point (–8,2) and has a slope of 1/2?
madam [21]
Y=1/2x+6

Y-2=1/2(x+8)
Then simplify and get y by itself
5 0
3 years ago
If 3x + 6 = 18, then x = ?
zaharov [31]

Answer:

X=4

Step-by-step explanation:

3x + 6 = 18 \\ 3x = 18 - 6 \\ 3x = 12 \\ x = 4

7 0
3 years ago
Please dont ignore, Need help!!! Use the law of sines/cosines to find..
Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

<h3>16</h3>

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

  • \sin{A} = \sin{103\textdegree{}},
  • The opposite side of angle A a = BC = 26,
  • The angle C is to be found, and
  • The length of the side opposite to angle C c = AB = 6.

\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}.

\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}.

\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}.

Note that the inverse sine function here \sin^{-1}() is also known as arcsin.

<h3>17</h3>

By the law of cosine,

c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C},

where

  • a, b, and c are the lengths of sides of triangle ABC, and
  • \cos{C} is the cosine of angle C.

For triangle ABC:

  • b = 21,
  • c = 30,
  • The length of a (segment BC) is to be found, and
  • The cosine of angle A is \cos{123\textdegree}.

Therefore, replace C in the equation with A, and the law of cosine will become:

a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}.

\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}.

<h3>18</h3>

For triangle ABC:

  • a = 14,
  • b = 9,
  • c = 6, and
  • Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}.

\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}.

\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree.

<h3>15</h3>

For triangle DEF:

  • The length of segment DF is to be found,
  • The length of segment EF is 9,
  • The sine of angle E is \sin{64\textdegree}}, and
  • The sine of angle D is \sin{39\textdegree}.

Apply the law of sine:

\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}

\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9.

7 0
3 years ago
Please help! Thank you!
KIM [24]
The correct answer is:  [B]:  " (2, 5) ".
__________________________________________
Given:
__________________________________________
  -5x + y = -5 ;
  -4x + 2y = 2 .
___________________________________________
  Consider the first equation:
___________________________
-5x + y = -5 ;  ↔ y + (-5x) = -5 ;

↔ y - 5x = -5 ;  Add "5x" to each side of the equation; to isolate "y" on one side of the equation; and to solve in terms of "y".
_____________________________________________
   y - 5x + 5x = -5 + 5x  

   y = -5 + 5x ;  ↔ y = 5x - 5 ;
____________________________________________
 Now, take our second equation:
______________________________
     -4x + 2y = 2 ; and plug in "(5x - 5)" for "y" ;  and solve for "x" :
_____________________________________________________
         -4x + 2(5x - 5) = 2 ; 
______________________________________________________
Note, 2(5x - 5) = 2(5x) - 2(5)  = 10x - 10 ;
__________________________________________
So:   -4x + 10x - 10 = 2 ;

On the left-hand side of the equation, combine the "like terms" ;

-4x +10x = 6x ;  and rewrite:

6x - 10 = 2  ;

Now, add "10" to each side of the equation:

6x - 10 + 10 = 2 + 10 ;

to get:

           6x = 12 ;  Now, divide EACH side of the equation by "6" ; to isolate "x" on one side of the equation; and to solve for "x" ; 

           6x/6 = 12 / 6 ;
 
                 x = 2 ;
_________________________________
Now, take our first given equation; and plug our solved value for "x" ; which is "2" ; and solve for "y" ;
_____________________________________
-5x + y = -5  ;

-5(2) + y = -5 ;

-10 + y = -5 ;  ↔
                              y - 10 = -5  ;

Add "10" to each side of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;
     
         y - 10 + 10 = -5 + 10 ;

                   y = 5 .
_____________________________
So, we have, x = 2 ; and y = 5 .
____________________________
Now, let us check our work by plugging in "2" for "x" and "5" for "y" in BOTH the original first and second equations:
______________________________
first equation: 

-5x + y = -5 ;

-5(2) + 5 =? -5?

-10 + 5 =? -5 ? YES!
______________________
second equation:

-4x + 2y = 2 ;

-4(2) + 2(5) =? 2 ?

-8 + 10 =? 2 ?  Yes!
_______________________________________________________
So, the answer is: 
___________________________________________________________
          x = 2 , y = 5 ; or, "(2, 5)" ;  which is: "Answer choice: [B] " .
___________________________________________________________



3 0
3 years ago
Are these two answers correct??
Damm [24]

Answer:

Yes

Step-by-step explanation:

7 0
2 years ago
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