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olga nikolaevna [1]
3 years ago
6

A candy store called "Sugar" built a giant hollow sugar cube out of wood to hang above the entrance to their store. It took 5454

5454 feet2^22start superscript, 2, end superscript of material to build the cube.
What is the volume inside the giant sugar cube?
Mathematics
1 answer:
Phoenix [80]3 years ago
4 0
The volume inside is 27 cu. ft.

The surface area of a cube is the area of each face of the cube added together.  There are 6 faces on a cube; 54/6 = 9 sq. ft. for each face of the cube.

Each face of a cube is a square, whose area is given by A=s²:
9 = s²

Take the square root of both sides:
√9 = √s²
3 = s

Each side of the square, or edge of the cube, is 3 ft.

This makes the volume 3(3)(3) = 27 cu. ft.
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(2,12)  , (4,7)  -> coefficient is negative, because the x-coordinate is increasing and y-coordinate is decreasing

(4,7),  (5,4)  ->  coeficien is negative too - look above
8 0
3 years ago
Solve the equation Rx+qx-d=gc
guajiro [1.7K]
Solve for x?
x(R + q) - d = gc
x(R = q) = gc + d
x = \frac{gc+d}{R + q}


4 0
3 years ago
Which expression is equivalent to 54n-20m+6n
Eva8 [605]

Answer:

C. 594 > -11n

Step-by-step explanation:

i took the test on edgen

4 0
3 years ago
Read 2 more answers
Does anyone know this ? i'm stuck.. i'll give you brainliest and 15 points !! show work.
madam [21]

Answer:

x = 7

Step-by-step explanation:

The angle adjacent to the 70 degree angle is also 70 degrees.  We assume that there are four 90-degree angles in the center of the figure.  Thus, 8x - 36 + 70 + 90 = 180 (the sum of the interior angles is 180 degrees).

Solving  8x - 36 + 70 + 90 = 180 for x:

              8x - 36 = 20, or 8x = 56.

Dividing both sides by 8 yields x = 7.

5 0
3 years ago
Please help me on this problem
Anna71 [15]

Answer:

The pairs of integer having two real solution forax^{2} -6x+c = 0 are

  1. a = -4, c = 5
  2. a = 1, c = 6
  3. a = 2, c = 3
  4. a = 3, c = 3

Step-by-step explanation:

Given

ax^{2} -6x+c = 0

Now we will solve the equation by putting all the 6 pairs so we get the  following

-3x^{2} -6x-5 = 0 for a = -3 , c=-5

-4x^{2} -6x+5 = 0 for a = -4 , c=5

1x^{2} -6x+6 = 0 for a = 1 , c=6

2x^{2} -6x+3 = 0 for a = 2 , c=3

3x^{2} -6x+3 = 0 for a = 3 , c=3

5x^{2} -6x+4 = 0 for a = 5 , c=4

The above  all are Quadratic equations inn general form ax^{2} +bx+c=0

where we have a,b and c constant values

So for a real Solution we must have

Disciminant , b^{2} -4\timesa\timesc \geq 0

for a = -3 , c=-5 we have

Discriminant =-24 which is less than 0 ∴ not a real solution.

for a = -4 , c=5 we have

Discriminant = 116 which is greater than 0 ∴ a real solution.

for a = 1 , c=6 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 2 , c=3 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 3 , c=3 we have

Discriminant =0 which is equal to 0 ∴ a real solution.

for a = 5 , c=4 we have

Discriminant =-44 which is less than 0 ∴ not a real solution.

7 0
3 years ago
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