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NemiM [27]
3 years ago
15

Find the side of a square, whose area is equal to the area of a rectangle with sides 6.4m and 2.5m. Also find the perimeter of t

he square.
Mathematics
1 answer:
san4es73 [151]3 years ago
5 0

Answer:

side: 4 metres

perimeter: 16 metres

Step-by-step explanation:

Let's first find the area of this rectangle.

The area of a rectangle is denoted by A = lw, where l is the length and w is the width. Here, the length is l = 6.4 and the width is w = 2.5. Plug these in:

A = lw

A = 6.4 * 2.5 = 16 metres squared

We want to find the side of a square with area 16. Suppose the side length is x. The area of a square is denoted by A = x * x = x², so set this equal to 16:

x² = 16

x = √16 = 4

Thus, the side length is 4 metres.

The perimeter of a square is denoted by P = 4s, where s is the side length.

Here the side length is 4 metres, as we found, so:

P = 4s = 4 * 4 = 16

Hence the perimeter is 16 metres.

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Find the measure of < AED for m < BEC = 96
Strike441 [17]
I wish you would of put a picture but with what you gave me. if the angle is a straight line then you would just set it up like 96+x=180 bc a straight line ='s 180 and den subtract 96 from 180 which is 84 so x would = 84 if the angle is a supp. angle but til there's not a picture this might be wrong.
But i hope this helps have a nice nite!
5 0
3 years ago
Read 2 more answers
WILL GIVE BRANLIEST!!! Pls help! Determine the coordinates of the point on the straight line y=3x+1 that is equidistant from the
iren [92.7K]

Let , coordinate of points are P( h,k ).

Also , k = 3h + 1

Distance of P from origin :

d=\sqrt{h^2+k^2}

Distance of P from ( -3, 4 ) :

d=\sqrt{(h+3)^2+(k-4)^2}

Now , these distance are equal :

h^2+(3h+1)^2=(h+3)^2+(3h+1-4)^2\\\\h^2+(3h+1)^2=(h+3)^2+(3h-3)^2

Solving above equation , we get :

P=(\dfrac{16}{21},\dfrac{23}{7})

Hence , this is the required solution.

6 0
3 years ago
The graph shows the feasible region for the system with constraints:
Dimas [21]

Answer:

The vertices feasible region are (0 , 15) , (10 , 15) , (20 , 5)

The minimum value of the objective function C is 125

Step-by-step explanation:

* Lets look to the graph to answer the question

- There are 3 inequalities

# y ≤ 15 represented by horizontal line (purple line) and cut the

  y-axis at point (0 , 15)

# x + y ≤ 25 represented by a line (green line) and intersected the

  x-axis at point (25 , 0) and the y- axis at point (0 , 25)

# x + 2y ≥ 30 represented by a line (blue line) and intersected the

  x-axis at point (30 , 0) and the y-axis at point (0 , 15)

- The three lines intersect each other in three points

# The blue and purple lines intersected in point (0 , 15)

# The green and the purple lines intersected in point (10 , 15)

# The green and the blue lines intersected in point (20 , 5)

- The three lines bounded the feasible region

∴ The vertices feasible region are (0 , 15) , (10 , 15) , (20 , 5)

- To find the minimum value of the objective function C = 4x + 9y,

  substitute the three vertices of the feasible region in C and chose

  the least answer

∵ C = 4x + 9y

- Use point (0 , 15)

∴ C = 4(0) + 9(15) = 0 + 135 = 135

- Use point (10 , 15)

∴ C = 4(10) + 9(15) = 40 + 135 = 175

- Use point (20 , 5)

∴ C = 4(40) + 9(5) = 80 + 45 = 125

- From all answers the least value is 125

∴ The minimum value of the objective function C is 125

6 0
3 years ago
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Express the set of real numbers between but not including 2 and 5 as follows.
Mashcka [7]
2 < x < 5

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Plzzz Help me Fast!!
taurus [48]

Answer:

Is each unit a cm

Step-by-step explanation:

3 0
3 years ago
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