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jekas [21]
2 years ago
7

Four Boys and Five Girls went to a water park. The total cost for the group was $144. What was the cost of admission for each pe

rson?
Mathematics
2 answers:
igomit [66]2 years ago
4 0

Answer:

16

Step-by-step explanation:

9 divided by 144 gives you 16

Goshia [24]2 years ago
3 0
The answer is 16 each person
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EXERCISE 1.2
natta225 [31]

sorry di pa namin yan napapag aralan..

4 0
3 years ago
Read 2 more answers
2. Tina uSUally brings her children every Sunday afternoon at the children's park in their barangay. The park is 350m Iong and 2
Verdich [7]

Answer: kindly check explanation

Step-by-step explanation:

Given the following :

Total area of parknand path way = 74,464 m2.

Dimension of park = 350m by 200m

Using the area of rectangle formula:

Area = Length * width

Area of park = 350m * 200m

Area of park = 70,000

Area of path = (74,464 - 70,000) = 4,464m2

Since path has uniform width, then width of the the path is sqrt(4464) = 66.8m

2.) the path of the park seems a bit narrow, hence, I wouldn't love children to bike and jog in such a narrow path

4 0
3 years ago
During the month of February, fabulous feet shoe mart sold 48 pairs of red loafers. After an ad campaign to boost sales, they so
Anvisha [2.4K]
<span>Sales went from 48 pairs to 60 pairs: total change = 60 - 48 = 12 percentage change = 12 / 48 = 0.25 = 0.25*100% = 25% Answer: 25%</span>
8 0
3 years ago
Read 2 more answers
What is the greatest common factor of 60x^4, 45x^5y^5 and 75x^3y
OLEGan [10]
Factor each
60x^4=2*2*3*5*x*x*x*x
45x^5y^5=3*3*5*x*x*x*x*x*y*y*y*y*y
75x^3y=3*5*5*x*x*x*y

common is 3*5*x*x*x=15x^3

gcf=15x^3
8 0
3 years ago
So the surface area is for the figure itself, and I just need to find the height. Please help!! (Geometry problem)
Harman [31]
We need to cut it up and find the areas of each
ok so base=a^2
back sides are a^2 and a^2
total we have a^2+a^2+a^2=3a^2

we have now the 4 triangles, 2 of which are identical
the  identical triangles are the side ones which are base a and height 1/2a
area of triangle=1/2bh
so the areas of them are 1/4a^2, times 2 since 2 of those triangles, 1/2a^2

add
3a^2+1/2a^2=7/4a^2

now the very top triangle and the cut corner triangle

very top is legnth a and base a so 1/2 a^2 is area
add
3 and 1/2a^2+1/2a^2=4a^2

we also need the small rectangles on side which is 1/2a times a times 2 since 2 of them=a^2 so a^2+4a^2=5a^2
now to find the corner bit in terms of a
use some imagination and lines to find legnths, right triangles and hypotonuse
the very top side is a√2 (45,45,90 angle/triangle)
we need to find slant height which is (a√6)/2

now
area=1/2bh
area=((a^2)√3)

add that
5a^2+(a^2)√3=250.5
find a
undistribute the a^2
(a^2)(5+√3)=501
use divide both sides by (5+√3)
a^2=501/(5+√3)
take squaer root of both sides
a= aprox 8.6267094364168 in

I hope this is right (should be, but alot done in head so maybe incorrect or off)

8 0
3 years ago
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