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aleksandrvk [35]
2 years ago
14

Denis made a scale drawing of his backyard, using the scale 1/4 inch = 3 feet. The rectangular swimming pool was 2 inches long a

nd 1 inch wide in the drawing. What was the area of the actual swimming pool? Show all the steps!
Mathematics
2 answers:
NeTakaya2 years ago
4 0
2 / .25 = 8; 8 * 3 = 24 ft
1 / .25 = 4; 4 * 3 = 12 ft
12 * 24 = 288 ft^2


Vlad [161]2 years ago
3 0
24ft long, 12ft wide

2/.25=8  8x3=24
1/.25=4  4x3=12
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Tracy had a bag of candies, and none of the candies could be broken into pieces. She ate $\frac{1}{3}$ of them and then gave $\f
Ymorist [56]

Answer:

72

Step-by-step explanation:

Let the number of candies = n

Tracy ate = 1/3 of n = n/3 remaining 2n / 3

she gave 1/4 of the remainder to her friend = 1/4 of ( n - n/3) = 2n / 12

the new remainder = (2n / 3) - (2n / 12) = n /2

she and her mom then ate altogether = 30

and the brother took from 1 to five, the number he took = x

and she has 3 left

( n/2) - 30 - x = 3

n = 2 ( 33 + x)

now we know that the candies could not be broken  and x is between 1 to 5

n = 2 (33 + 1), 2 (33+2), 2(33 +3), 2 (33+ 4) and 2 (33 + 5) this are the possible values of n, ( 68, 70, 72, 74, 76) and the multiple of 3 is 72

n therefore = 72

7 0
3 years ago
X/3+9=17 simplify your answer
frosja888 [35]
I think the answer might be x=24
8 0
2 years ago
Write and graph a system of inequalities that models the situation.
liubo4ka [24]

Answer:

The feasible region in the attached figure

Step-by-step explanation:

Let

x ----> the number of shirts

y ----> the number of pants

we know that

The cost of the shirts (number of shirts multiplied by the cost of one shirt) plus the cost of the pants (number of pants multiplied by the cost of one pant) must be less than or equal to $80

so

The inequality that represent this problem is

5x+8y\leq 80

using a graphing tool

The feasible region is the triangular shaded area

see the attached figure

7 0
3 years ago
(1 ÷ x- 1)+(2÷ x+2)=(3÷2) solve and check for extraneous solutions
Sliva [168]
\frac{1}{x-1}+ \frac{2}{x+2}= \frac{3}{2}
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\frac{x+2+2(x-1)}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}
\frac{x+2+2x-1}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}
\frac{3x+1}{(x-1)(x+2)}= \frac{3(x^2+x-2)}{2}
\frac{}{(x-1)(x+2)}= \frac{3x^2+3x-6)}{2}
\frac{3x+1}{(x^2+x-2)}= \frac{3x^2+3x-6)}{2}
\frac{2(3x+1)}{2(x^2+x-2)}= \frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}
\frac{2(3x+1)}{2(x^2+x-2)}-\frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}=0
\frac{2(3x+1)-3x^2+3x-6}{2(x^2+x-2)}=0

this will be continued
5 0
3 years ago
USA test prep please answer
sergij07 [2.7K]

Answer:

D

Step-by-step explanation:

8 0
2 years ago
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