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Eddi Din [679]
2 years ago
6

-2-5(4n-6)=-132 not really sure how to do this one ..

Mathematics
2 answers:
uranmaximum [27]2 years ago
8 0

All you have to do is follow PEMDAS (if you don't know what that is, search it up).


-2-5(4n-6)=-132

Distribute the 5 to everything in the parentheses.

-2-20n+30= -132

Combine like terms.

-20n+28 = -132

Subtract 28 from both sides.

-20n = -160

Divide by -20 on both sides.

n = 8

Nikolay [14]2 years ago
8 0

Use order of operations rules to evaluate all terms of -2-5(4n-6)=-132, and then solve for n:

Must multiply -5(4n-6) first: -20n + 30. Then we have -2 - 20n + 30 = -132.

Adding 2 to both sides, we get -20n + 30 = -130, or -20n = -160.

Solving for n, n = 160/20 = 8 = n

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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify
irina [24]

Answer:

x-coordinates of relative extrema = \frac{-1}{2}

x-coordinates of the inflexion points are 0, 1

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f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}

Differentiate with respect to x

f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}

f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}

Differentiate f'(x) with respect to x

f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1

At x = \frac{-1}{2},

f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0

We know that if f''(a)>0 then x = a is a point of minima.

So, x=\frac{-1}{2} is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0
3 years ago
3(0.3x +1.3) = 2(0.4x -0.85)
Verdich [7]

Answer:

Step-by-step explanation:

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0.9x + 3.9 = 0.8x - 1.70

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5 0
2 years ago
Help me solve this please
shutvik [7]

Answer:

D'G' = 52.5 units

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Since the dilatation is centred at the origin then multiply the original coordinates by the scale factor 3.5

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G' = (- 8 × 3.5, - 5 × 3.5 ) = (- 28, - 17.5 )

Calculate D'G' using the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = D' (3.5, 24.5) and (x₂, y₂ ) = G' (- 28, - 17.5)

D'G' = \sqrt{(-28-3.5)^2+(-17.5-24.5)^2}

       = \sqrt{(-31.5)^2+42^2}

       = \sqrt{992.25+1764}

        = \sqrt{2756.25}

         = 52.5 units

7 0
2 years ago
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