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Thepotemich [5.8K]
2 years ago
8

How many squares are in the 11 squares

Mathematics
2 answers:
Nadya [2.5K]2 years ago
7 0

Answer:

11 Squares are there in 11 squares

motikmotik2 years ago
6 0

Answer:

11

Step-by-step explanation:

because take your question and realize you put, "How many squares are in the 11 squares?"

(Sry if wrong or information misunderstood!)

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(solve the following problem using integers. show ALL your work.)
yulyashka [42]

Answer:

$4

Step-by-step explanation:

30+20=50

50-48=2

2+20+=22

22-18=4

if she has 30 then gets 20 she has 50. Then she spends 48 dollars leaving her with only 2 dollars. She then makes another 20 dollars so she now has 22 dollars. She spends 18 dollars leaving her with 4 dollars.

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3 years ago
To control pollination, pollen-producing flowers are often removed from the top of corn in a process called detasseling. The hou
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3 years ago
Jerry runs up to a tank that holds up to 15 gallons of melted cheese When it is active it drains cheese onto crackers at a const
frosja888 [35]

Answer:

.6 gallons per minute

Step-by-step explanation:

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8 0
3 years ago
9+10 ii am struggling really bad plz help.
Valentin [98]

Answer:


Step-by-step explanation:

9+10=19.

Replace the 0 in the ten with the 9.


3 0
3 years ago
A piece of wire 23 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
AURORKA [14]

Answer:

For maximum area, all of the wire should be used to construct the square.

The minimum total area is obtained when length of the wire is 10m

Step-by-step explanation:

For maximum,  we use the whole length

For minimum,

supposed the x length was used for the square,

the length of the side of the square = x/4m

Area = \frac{x^{2} }{16}

For the equilateral triangle, the length of the side =  \frac{23 - x}{3}

Area = \frac{\sqrt{3} }{4}  a^{2}  = \frac{\sqrt{3} }{4} (\frac{23 - x}{3} )^{2}

Total Area = \frac{x^{2} }{16}  + \frac{\sqrt{3} }{36} (23-x)^{2}

\frac{dA}{dx}  = \frac{x}{8}  -  \frac{\sqrt{3} }{18} (23 - x)\\

\frac{d^{2}A }{dx^{2} }  = \frac{1}{8}  + \frac{\sqrt{3} }{18}  > 0, therefore it is minimum

\frac{dA}{dx}  = 0 \\\\

\frac{x}{8}  -  \frac{\sqrt{3} }{18} (23 - x) = 0\\

x = 10.00m

3 0
3 years ago
Read 2 more answers
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