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7nadin3 [17]
3 years ago
12

Nuts Galore sells a trail mix containing dried banana chips, roasted almonds, and chocolate chips that sells for $6.52 per pound

. The dried banana chips costs $4 per pound, the roasted almonds cost $8 per pound, and the chocolate chips cost $6 per pound. The amount of roasted almonds used is three times the amount of chocolate chips used. If they plan to make 100 pounds of this trail mix, how many pounds of each ingredient should be used
Mathematics
1 answer:
KengaRu [80]3 years ago
6 0

Answer:

x = 28 ----- Banana

y = 54 ----- Almonds

z = 18 ---- Chocolate

Step-by-step explanation:

Let

x \to banana

y \to almonds

z \to chocolate

So, we have:

x + y + z = 100 --- total used

4x + 8y + 6z = 652 -- i.e. 6.52 * 100 --- total sales from 100 pounds

y = 3z

Required

The number of pounds of each (i.e. x, y and z)

Substitute y = 3z in x + y + z = 100

x + 3z + z = 100

x + 4z = 100

Make x the subject

x = 100 - 4z

Substitute y = 3z in 4x + 8y + 6z = 652

4x + 8 * 3z + 6z = 652

4x + 24z + 6z = 652

4x + 30z = 652

Substitute x = 100 - 4z

4(100 - 4z) + 30z = 652

400 - 16z + 30z = 652

Collect like terms

- 16z + 30z = 652-400

14z = 252

Divide by 14

z = 18

Substitute z = 18 in x = 100 - 4z

x = 100 - 4 * 18

x = 100 - 72

x = 28

To calculate y, we have:

y = 3z

y = 3 * 18

y = 54

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1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.
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First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}

(If you were to plot the actual curve, you would have both y=x^{2/3} and y=-x^{2/3}, but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx

We have

y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}

Then in the integral,

\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx

Substitute

u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx

This transforms the integral to

\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du

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\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)

We can simplify this further to

\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}

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