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Archy [21]
3 years ago
6

What expression must the center cell of the table below contain so that the sums of each row, each column, and each diagonal are

equivalent?
  x       8x      -3x
-2x      ?        6x
7x      -4x       3x
Mathematics
1 answer:
sergey [27]3 years ago
5 0
Oh, I adore magic squares!

We need all rows to be equivalent, so try adding up the bottom (complete) row to get 7x-4x+3x=10x-4x=6x.

To solve the middle number, we have -2x+X+6x=6x, so X=2x.
To solve the top number, we have X+8x-3x=6x, so X=6x-8x+3x=x

Now check first column, 
x-2x+7x=6x   ok
Second column
8x+2x-4x=6x  ok
Top-right to bottom-right diagonal: x+2x+3x=6x, ok   
Bottom-left to top right diagonal: 7x+2x-3x=6x

So everything is ok!
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Explanation:

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3 0
3 years ago
Which expression is equivalent to 9q2-(3q - 7) +5q??<br> ) ?<br> 3
lesantik [10]

9q^2-(3q - 7) +5q= \\\\=9q^2-3q + 7 +5q= \\\\=9q^2+2q + 7

3 0
3 years ago
Find the number in the proportion <br> 2/12=5/x
Sliva [168]

Answer: x = 30

2/12 x 5/x

(Cross Multiply)

2x = 60

x = 30

3 0
3 years ago
Find the value of each variable in the parallelogram
wlad13 [49]

Answer:

B= 90

C= 80

D= 100

Step-by-step explanation:

So let’s start with the bottom angles.

We know that the adjacent angles in a parallelogram add up to 180° so our equation would be (b-10) + (b+10)=180. So let’s simplify it b and b are both the same variables so we will get 2b and -10 + 10 is 0 so we get the following equation 2b=180.

So now we divide 180 and 2 and we get 90 as b.

So one side will be 100 because of the +10 and the other 90 because of the -10.

Now for the next side.

d+c=180

So we know that b+10 is adjacent to c so 180-100 is 80, so 80 is c.

Now for d.

b-10 is 80 so 180-80 is 100, so d is 100.

7 0
3 years ago
Object is thrown upward from a height of 15 ft at an initial vertical velocity of 30 ft per second. How long will it take to hit
dem82 [27]

Answer:

2.25 s.

Step-by-step explanation:

We'll begin by calculating the time taken for the object to get to the maximum height from the point of projection. This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₁) to reach the maximum height from the point of projection =?

v = u – gt₁ (since the object is going against gravity)

0 = 30 – (32.15 × t₁

0 = 30 – 32.15t₁

Collect like terms

0 – 30 = – 32.15t₁

– 30 = – 32.15t₁

Divide both side by – 32.15

t₁ = –30 / –32.15

t₁ = 0.93 s

Next, we shall determine the maximum height reached by the object from the point of projection.

This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Maximum height (h) reached from the point of projection =?

v² = u² – 2gh (since the object is going against gravity)

0² = 30² – (2 × 32.15 × h)

0 = 900 – 64.3h

Collect like terms

0 – 900 = – 64.3h

– 900 = – 64.3h

Divide both side by – 64.3

h = –900 / –64.3

h = 14 ft

Thus, the maximum reached by the object from the point of projection is 14 ft.

Next, we shall determine the height to which the of object is located from the maximum height reached to the ground. This can be obtained as follow:

Height (h₀) from which the object was projected = 14 ft

Maximum Height (h) reached from the point of projection = 14 ft

Height (hₗ) to which the of object is located from the maximum to the ground =?

hₗ = h₀ + h

hₗ = 14 + 14

hₗ = 28 ft

Thus, the height to which the of object is located from the maximum reached to the ground is 28 ft.

Next, we shall determine the time taken for the object to get to the ground from the maximum height reached. This can be obtained as follow:

Height (hₗ) to which the of object is located from the maximum to the ground = 28 ft

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₂) taken for the object to get to the ground from the maximum height reached =?

hₗ = ½gt₂²

28 = ½ × 32.15 × t₂²

28 = 16.075 × t₂²

Divide both side by 16.075

t₂² = 28 / 16.075

Take the square root of both side

t₂ = √(28 / 16.075)

t₂ = 1.32 s

Finally, we shall determine the time take for the object to get to the ground from the point of projection. This can be obtained as follow:

Time (t₁) to reach the maximum height from the point of projection = 0.93 s

Time (t₂) taken for the object to get to the ground from the maximum height reached = 1.32 s

Time (T) take for the object to get to the ground from the point of projection =?

T = t₁ + t₂

T = 0.93 + 1.32

T = 2.25 s.

Therefore, the time take for the object to get to the ground from the point of projection is 2.25 s.

7 0
3 years ago
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