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Lapatulllka [165]
3 years ago
5

What is the area, in square feet, of the Free Play section of the children’s playroom?

Mathematics
1 answer:
Bumek [7]3 years ago
8 0

Answer:

90ft^{2}

Step-by-step explanation:

The width of the free play section is 9 feet. The length is 5+5, which equals 10.

The area is l×w.

9×10=90 feet square

<u>Hope this helps :-)</u>

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Sergeu [11.5K]

Answer:

m = 10

Step-by-step explanation:

Find 'n' first..... then pythag theorem for 'm'....

n is to 5    as  15 is to n

n/5   = 15/n

n^2 = 75     n = sqrt (75)

Then   m^2 =  n^2 + 5^2

            m^2 = 75  +  25       s-o-o-o-o:     m = 10

6 0
2 years ago
Find the values of y for which the distance between the point p2,-3 and q 10, y is 10 units
lina2011 [118]

Answer:

y = - 9, y = 3

Step-by-step explanation:

Calculate distance d using the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = (2, - 3) and (x₂, y₂ ) = (10, y)

d = \sqrt{(10-2)^2+(y+3)^2}

   = \sqrt{8^2+(y+3)^2}

Given distance between points is 10, then

\sqrt{64 +(y+3)^2} = 10 ( square both sides )

64 + (y + 3)² = 100 ( subtract 64 from both sides )

(y + 3)² = 36 ( take the square root of both sides )

y + 3 = ± \sqrt{36} = ± 6 ( subtract 3 from both sides )

y = - 3 ± 6 , thus

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y = - 3 + 6 = 3

7 0
3 years ago
3 + c = -14 what does c equal
vodka [1.7K]
The answer is 11 so C=11.

8 0
3 years ago
A proton moves at 5.20 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40
lozanna [386]

Answer:

  86.5 ns

Step-by-step explanation:

The speed in the original direction (horizontally) is unchanged by the vertical force the field exerts. The travel time is ...

  time = distance/speed = (4.5×10^-2 m)/(5.20×10^5 m/s) = 8.65×10^-8 s

_____

An engineer would express this time using the SI prefix nano- for 10^-9. The time is 86.5 ns.

7 0
3 years ago
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