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kondor19780726 [428]

2 years ago

3

marshall bought 32 ounces of mixed nuts, which are estimated to be 30%. which expression can be used to find the percentage of p

eanut concentration of the final mix if he adds x ounces of peanuts?

2 answers:

Goryan [66]2 years ago

3 0

<span>Given that Marshall bought 32 ounces of mixed nuts, which are estimated to be 30% peanuts. This means that the amount of peanut in the mixed nut is given by 0.3(32) = 9.6 ounces. Then x ounces of peanuts is added to the mixture, the new amount of peanut is 9.6 + x ounces and the new total amount of mixed nuts is 32 + x ounces. Therefore, the expression that can be used to find the percentage of peanut concentration of the final mix if he adds x ounces of peanuts is given by (9.6 + x) / (32 + x) x 100%</span>

ANEK [815]2 years ago

3 0

Answer: y=0.3(32)+x

------------- *100

32+x

Step-by-step explanation:

Answer on Edgen

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Answer:

The required equation is:

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Explanation:

Let us assume that the hole is at y = 0m, with x as the time.

From the question we have (-1s, 8m) as the vertex (here x being the time variable is supposed to be in seconds and y being the distance variable is supposed to be in meters)

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We know that the vertex of the parabola y = ax² + bx + c is at

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therefore we have:

$-1 = \frac{-b}{2a}$

We then have the following equations:

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From the 3rd equation we have

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Therefore we have:

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We can simplify both equations and get:

$8 = a\times( -1^2 - 2s^2) + c = -a\times 3^2 + c$

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With a, we can find the values of c and b.

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$y = -\frac{4}{3}t^2 -\frac{8}{3}t + 4$

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The complete question in the attached figure

we know that
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y=a(x-h)²+k
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so
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Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

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<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

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Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

$g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9$

Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

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Checking for inverse functions

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Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

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Divide through by 3

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$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

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