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arlik [135]
3 years ago
7

Please help, you get 20 points for this!

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

2.15 times or 215%

Step-by-step explanation:

V(Luxor) = (646)² · 350/3

V(Tower) = 290 x 125 x 622

V(Luxor) ÷ V(Tower) = 2.15

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A complete table with values for x or y that make this equation true:3x+y=15
Andreyy89

Answer:


Step-by-step explanation:

The given equation is: 3x+y=15,

Now, putting x=2 in above equation,

3(2)+y=15

y=15-6=9

Now, put y=3 in the given equation,

3x+3=15

3x=12

x=4

Putting x=6,

3(6)+y=15

y=15-18y=-3

Putting x=0,

0+y=15

y=15

Putting x=3,

3(3)+y=15

y=15-9

y=6

Putting y=0,

3x+0=15

x=5

Putting y=8,

3x+8=15

3x=7

x=\frac{7}{3}

4 0
3 years ago
Find the domain for the particular solution to the differential equation dy dx equals the quotient of 3 times y and x , with ini
Svetach [21]
For this case we have the following difference equation:
 dy / dx = 3xy

 Applying separable variables we have:
 dy / y = 3xdx

 Integrating both sides we have:
 \int\ ({1/y}) \, dy =  \int\ {3x} \, dx
 ln (y) = (3/2) x ^ 2 + C

 applying exponential to both sides:
 exp (ln (y)) = exp ((3/2) x ^ 2 + C)

 y = C * exp ((3/2) x ^ 2)
 For y (1) = 1 we have:
 C = 1 / (exp ((3/2) * 1 ^ 2))

C = 0.2
 Thus, the particular solution is:
 y = 0.2 * exp ((3/2) x ^ 2)

 Whose domain is all real.
 Answer:
 y = 0.2 * exp ((3/2) x ^ 2)
 Domain: all real numbers
6 0
3 years ago
I need help with this
pogonyaev

Answer:

no solutions

Step-by-step explanation:

3x+ 7(x+1) = 2(6x+5) -2x

Distribute

3x+7x+7 = 12x+10-2x

Combine like terms

10x+7 = 10x+10

Subtract 10 from each side

10x+7-10x = 10x+10-10x

7 = 10

This is not a true statement so there are no solutions

4 0
3 years ago
(12,-1), (-2,-3)<br>please help​
Leviafan [203]

Answer:

13, _1 (12) (-2,3) 1 answer ...<em>by</em><em> </em><em> </em><em>maths</em><em> </em>

4 0
2 years ago
How many times can 2500 go into 25
Alex_Xolod [135]

Answer:

100 times

25×100=2500

FOLLOW ME FOR CLEARING YOUR NEXT DOUBT

4 0
3 years ago
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