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3 years ago

15

Help me simplify this, please!!

1 answer:

OlgaM077 [116]3 years ago

6 0

$- \sqrt[3]{2 x^{4} y^{2} } *3 \sqrt[3]{20 x^{5}y } =-3 \sqrt[3]{2 x^{4} y^{2}20 x^{5}y} = \\ \\ =-3 \sqrt[3]{40 x^{9} y^{3} } =-3 \sqrt[3]{2 ^{3}*5* x^{9} y^{3} } = \\ \\ =-3*2 ^{ \frac{3}{3} } * x^{ \frac{9}{3} } * y^{ \frac{3}{3} } \sqrt[3]{5} =-6 x^{3} y \sqrt[3]{5}$
Answer: B.

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Answer:

$(a)$ $V = -H^3+ 5H^2 + 36H$

$(b)$ $Length = 5in; Width = 9in; Height = 5in$ or $Length = 3in; Width = 10in; Height = 6in$

Step-by-step explanation:

Let:

$W=Width; L = Length; H = Height$

Given

$W = 4 + H$

$L = 9 - H$

Solving (a): The polynomial for the volume in terms of height

Volume (V) is calculated using:

$V = L * W * H$

$V = (9 - H) * (4 + H) * H$

Expand

$V = (9 - H) * (4H + H^2)$

Open brackets

$V = 36H + 9H^2 - 4H^2 - H^3$

$V = 36H + 5H^2 - H^3$

Rewrite as:

$V = -H^3+ 5H^2 + 36H$

Solving (b): The dimensions when Volume = 180

Substitute 180 for V in: $V = -H^3+ 5H^2 + 36H$

$180 = -H^3+ 5H^2 + 36H$

Rewrite as:

$H^3 - 5H^2 - 36H + 180 = 0$

Factorize:

$H^2(H - 5) - 36(H - 5) = 0$

$(H^2 - 36)(H - 5) = 0$

Express $H^2 - 36$ as difference of two squares

$(H - 6)(H + 6)(H - 5) =0$

This gives:

$H - 6 =0==> H = 6$

$H + 6 =0==> H = -6$

$H - 5 =0==> H = 5$

Since the height can not be negative, then H = 5 or H = 6

Recall that:

$W = 4 + H$

$L = 9 - H$

When H = 5

$W = 4 + 5 = 9$

$L = 9 - 5 = 4$

When H = 6

$W = 4 + 6 = 10$

$L = 9 - 6 = 3$

Hence, the dimensions of the box are:

$Length = 5in; Width = 9in; Height = 5in$

or

$Length = 3in; Width = 10in; Height = 6in$

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