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diamong [38]
3 years ago
12

There are 220 tickets to a school's annual awards banquet. Each adult ticket, a, cost $25 in each student ticket, s, cost $15 ev

ery ticket is sold in the school makes $4,650. Which system of linear equations could be used to find the number of adult and student tickets sold?
Mathematics
1 answer:
aleksley [76]3 years ago
7 0

Answer:

x + y = 220

25 x + 15 y =  4650

The above system can be used to find the number of adult and student tickets sold.

Step-by-step explanation:

Let us assume the number of Adult's ticket sold = x

And The number of student's tickets sold = y

Now, total number of tickets sold = 220

⇒ x + y = 220

Cost of 1 adult Ticket  = $ 25

⇒ Cost of x adult tickets =  x ( $25) = 25 x

Cost of 1 student Ticket  = $ 15

⇒ Cost of y student tickets =  y ( $15) = 15 x

Also, the total amount made from selling all tickets = $4650

⇒ 25 x + 15 y =  $4650

Here, the given system of equations are:

x + y = 220

and 25 x + 15 y =  4650

The above system can be used to find the number of adult and student tickets sold.

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Dmitry [639]

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h) Combine 5+4 to get 9. Take the root of 36, leaving you with 18 + 6.  Combine 18 + 6 to get 24. The answer is 24.


5. [15 + 22 + 53] divided by [12 + 18] = [90] divided by [30] = 3 ribbons each.

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4 0
3 years ago
If you add 12 marshmallows to each of 5 bags of marshmallows, and each bag started with the same number of marshmallows, the tot
ss7ja [257]
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5 0
3 years ago
HELP ASAP PLZ :)
Natalija [7]

for the triangle...

b=2h

a=(1\2).bh=648

(1/2).2h.h=648

h2=648

h-18|2 in

h=36|2 in

for the rectangle...

L=3+w

a=l+w=648

{3+w}w=648

w2+3w-648=0

Only positive W values make sense here..

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3 0
3 years ago
In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths. AB=4 and BC=3, Find AH, CH and BH.
IRISSAK [1]

Answer:

  • AH = 3.2
  • CH = 1.8
  • BH = 2.4

Step-by-step explanation:  

It can be convenient to compute the length of the hypotenuse of this triangle (AC). The Pythagorean theorem tells you ...

   AC^2 = AB^2 + CB^2

   AC^2 = 4^2 + 3^2 = 16 + 9 = 25

   AC = √25 = 5  

The altitude divides ∆ABC into similar triangles ∆AHB and ∆BHC. The scale factor for ∆AHB is ...

  scale factor ∆ABC to ∆AHB = AB/AC = 4/5 = 0.8  

And the scale factor to ∆BHC is ...  

  scale factor ∆ABC to ∆BHC = BC/AC = 3/5 = 0.6  

Then the side AH is 0.8·AB = 0.8·4 = 3.2

And the side CH is 0.6·BC = 0.6·3 = 1.8  

These two side lengths should add to the length AC = 5, and they do.  

The remaining side BH can be found from either scale factor:    

  BH = AB·0.6 = BC·0.8 = 4·0.6 = 3·0.8 = 2.4

_____  

The sides of interest are ...  

  AH = 3.2

  CH = 1.8

  BH = 2.4

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