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inysia [295]
3 years ago
14

Solve 12x+6>9x+12 please​

Mathematics
2 answers:
Inga [223]3 years ago
7 0

Answer:

x > 2 is your answer.

Step-by-step explanation:

Isolate the variable, x. Treat the > sign as an equal sign, what you do to one side, you do to the other.

First, subtract 9x & 6 from both sides.

12x (-9x) + 6 (-6) > 9x (-9x) + 12 (-6)

12x - 9x > 12 - 6

Simplify.

12x - 9x > 12 - 6

3x > 6

Isolate the variable (x). Divide 3 from both sides.

(3x)/3 > (6)/3

x > 6/3

x > 2

x > 2 is your answer.

~

alukav5142 [94]3 years ago
4 0

12x+6 > 9x+12

Subtract 6 from each side:

12x > 9x + 6

Subtract 9x from each side:

3x > 6

Divide both sides by 3:

x > 6 /3

X > 2

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vagabundo [1.1K]

Answer:

Move the variable term to the right hand side of the equation

Step-by-step explanation:

3 0
3 years ago
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Use the theorem in Sec. 28 to show that if f (z) is analytic and not constant throughout a domain D, then it cannot be constant
alexandr1967 [171]

Answer:

The value of f(z) is not constant in any neighbourhood of D. The proof is as explained in the explaination.

Step-by-step explanation:

Given

For any given function f(z), it is analytic and not constant throughout a domain D

To Prove

The function f(z) is non-constant constant in the neighbourhood lying in D.

Proof

1-Assume that the value of f(z)  is analytic and has a constant throughout some neighbourhood in D which is ω₀

2-Now consider another function F₁(z) where

F₁(z)=f(z)-ω₀

3-As f(z) is analytic throughout D and F₁(z) is a difference of an analytic function and a constant so it is also an analytic function.

4-Assume that the value of F₁(z) is 0 throughout the domain D thus F₁(z)≡0 in domain D.

5-Replacing value of F₁(z) in the above gives:

F₁(z)≡0 in domain D

f(z)-ω₀≡0 in domain D

f(z)≡0+ω₀ in domain D

f(z)≡ω₀ in domain D

So this indicates that the value of f(z) for all values in domain D is a constant  ω₀.

This contradicts with the initial given statement, where the value of f(z) is not constant thus the assumption is wrong and the value of f(z) is not constant in any neighbourhood of D.

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2 years ago
Write this number in standard form. 300+80+0.9+0.06+0.001
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Hey there! I'm happy to help!

First, let's add the hundreds and the tens.

300+80=380

We see that there is nothing in the ones place, so we keep our ones place 0 and we move onto adding the tenths.

380+0.9=380.9

We add the hundredths.

380.9+0.06=380.96

And finally, we add the thousandths.

380.96+0.001=380.961

Therefore, this number in standard form is 380.961.

Have a wonderful day! :D

5 0
2 years ago
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I need help on this loll
Serhud [2]

Answer: The translation rule that maps point D ( 7 , − 3 ) onto point D ' ( 2 , 5 )

is (x , y) → (x - 5 , y + 8)

Step-by-step explanation:

Let us revise the translation

If the point (x , y) translated horizontally to the right by h units  then its image is (x + h , y)

If the point (x , y) translated horizontally to the left by h units  then its image is (x - h , y)

If the point (x , y) translated vertically up by k units  then its image is (x , y + k)

If the point (x , y) translated vertically down by k units  then its image is (x , y - k)

(x , y) → (x ± h , y ± k) the right arrow symbol used to show the

translation from a point to its image

Example:

∵ P (0 , 0) → P' (1 , 2)

∴ The rule is (x , y) → (x + 1 , y + 2)

Let us find the translation rule that maps point D ( 7 , − 3 ) onto

point D' (2 , 5)

∵ Point (x , y) = (7 , -3)

∵ Its image after translation (x + h , y + k) = (2 , 5)

∴ x + h = 2

∵ x = 7

∴ 7 + h = 2

- Subtract 7 from both sides

∴ h = -5

∵ y + k = 5

∵ y = -3

∴ -3 + k = 5

- Add 3 to both sides

∴ k = 8

∴ The rule of translation is (x , y) → (x - 5 , y + 8)

The translation rule that maps point D ( 7 , − 3 ) onto point D ' ( 2 , 5 )

is (x , y) → (x - 5 , y + 8)

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