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STALIN [3.7K]
3 years ago
9

Please awnser questions, no links, no I dont know

Mathematics
2 answers:
german3 years ago
5 0
Use a calculator for dimensions that’s what I use just know which one is each side
Mrac [35]3 years ago
3 0

Answer:

330cm

Step-by-step explanation:

The formula to find the surface area of a rectangular prism is 2(lh+hw+lw)

(lh) in this formula is length times height

(hw) is height times width

(lw) is length times width

To find the answer to this problem lets make the height 3/5cm the width 10cm and the length 15cm

so now you would do (15*3/5) which is 9. This solve (lh)

then you would find (3/5*10) which is 6. This solves (hw)

and lastly you would find (15*10) which is 150. This solves (lw)

Now that you now the value of each one of these you replace it in the equation as so: 2(9+6+150)

now you solve the brackets which leaves us with 2(165)

now you multiply 2 with 165

and the answer is 330cm

Surface area = 330cm

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Mademuasel [1]

Answer:

{x = 1 , y=1, z=0

Step-by-step explanation:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

{-2 x - y + z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x - 3 y - 2 z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y+0 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer:  {x = 1 , y=1, z=0

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BARSIC [14]
1. Domain.

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So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.

2. Asymptotes:

f(x)=\dfrac{x-3}{x^2-2x-3}=\dfrac{x-3}{(x-3)(x+1)}=\boxed{\dfrac{1}{x+1}}

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