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sesenic [268]
3 years ago
12

Help me find my answer please

Mathematics
2 answers:
jonny [76]3 years ago
7 0

Answer:

-3

Step-by-step explanation:

\frac{-1}{3} =\frac{-3}{9}

Hunter-Best [27]3 years ago
3 0

Answer:

y = - 3

Step-by-step explanation:

Given that y varies directly with x then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition that y = - 1 when x = 3, thus

- 1 = 3k ( divide both sides by 3 )

- \frac{1}{3} = k

y = - \frac{1}{3} x ← equation of variation

When x = 9, then

y = - \frac{1}{3} × 9 = - 3

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Step-by-step explanation:

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2 years ago
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Lostsunrise [7]

Answer:

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3 0
2 years ago
7. y-2 = (x+3)<br> 4<br> 2<br> X<br> - TO<br> 12<br> 4
drek231 [11]

Answer:

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Step-by-step explanation:

Simplifying:

y = 2(x + 3)(x + -4)

reorder the terms:

y = 2(3 + x)(x + -4)

multiply (3 + x) * (-4 + x)

y = 2(3(-4 + x) + x(-4 + x))

y = 2((-4 * 3 + x * 3) + x(-4 + x))

y = 2((-12 + 3x) + x(-4 + x))

y = 2(-12 + 3x + (-4 * x + x * x))

y = 2(-12 + 3x + (-4x + x2))

Combine like terms  3x + -4x = -1x

y = 2(-12 + -1x + x2)

y = (-12 * 2 + -1x * 2 + x2 * 2)

y = (-24 + -2x + 2x2)

Solving:

y = -24 + -2x + 2x2

solving for variable 'y'

Move all terms containing y to the left, all other terms to the right.

Simplifying:

y = -24 + -2x + 2x2

8 0
3 years ago
Prove if the sum of the digits of a 3 digit number n is divisible by 9, then n is divisible by 9.
beks73 [17]

n=100x+10y+z\\x+y+z=9k,k\in\mathbb{Z}\\\\n=99x+9y+x+y+z\\n=99x+9y+9k\\n=9(11x+y+k)

6 0
3 years ago
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