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3 years ago

Please can I have some help thank you :)

Mathematics

1 answer:

Evgen [1.6K]3 years ago

8 0

9514 1404 393

Answer:

£592 for 16 boxes

Step-by-step explanation:

The classroom floor area is ...

A = LW = (8.8 m)(2.2 m) = 19.36 m²

The cupboard floor area is ...

A = (2.3 m)(2.1 m) = 4.83 m²

Then the total area to be covered is ...

19.36 m² +4.83 m² = 24.19 m²

At 1.6 m² per box, the number of boxes required is ...

(24.19 m²)/(1.6 m²/box) = 15.12 boxes

If we assume we can only purchase whole boxes, we need to purchase 16 boxes to cover the floor. The cost will be ...

(16 boxes)(£37/box) = £592 . . . . cost of laminate to replace the floor

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Let f (x, y) = x3 + y3 x2 + y2 . (a) Show that |x3|≤|x|(x2 + y2), |y3|≤|y|(x2 + y2) (b) Show that |f (x, y)|≤|x|+|y|. (c) Use th

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Answer and Step-by-step explanation:

a.)x^2 and y^2 are always greater than zero.(assuming real x and y).

=>x^2<=x^2+y^2

=>|x|(x^2)<=|x|(x^2+y^2)

=>|x^3|<=|x|(x^2+y^2)

b. similarly, |y^3|<=|y|(x^2+y^2)

=>|f(x,y)|=|x^3+y^3|/(x^2+y^2)=|x|+|y|

c. let h is very small number (close to zero but positive)

lim (x,y)-> (0,0) f(x,y)=(x^3+y^3)/(x^2+y^2)

putting x=h and y=h and approaching h->0

lim h->0 f(h)=2h=0

There is another explanation attached below

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3 years ago

How do you solve this? Thank you

V125BC [204]

2)

a)

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
(4x^5\cdot x^{\frac{1}{3}})+(2x^4\cdot x^{\frac{1}{3}})-(7x^3\cdot x^{\frac{1}{3}})+(3x^2\cdot x^{\frac{1}{3}})\\\\+(9x^1\cdot x^{\frac{1}{3}})-(1\cdot x^{\frac{1}{3}})
\\\\\\
4x^{5+\frac{1}{3}}+2x^{4+\frac{1}{3}}-7x^{3+\frac{1}{3}}+9x^{1+\frac{1}{3}}-x^{\frac{1}{3}}$

$\bf 4x^{\frac{16}{3}}+2x^{\frac{13}{3}}-7x^{\frac{10}{3}}+9x^{\frac{4}{3}}-x^{\frac{1}{3}}
\\\\\\
4\sqrt[3]{x^{16}}+2\sqrt[3]{x^{13}}-7\sqrt[3]{x^{10}}+9\sqrt[3]{x^4}-\sqrt[3]{x}$

b)

$\bf \cfrac{4x^5+2x^4-7x^3+3x^2+9x-1}{x^{\frac{1}{3}}}\impliedby \textit{distributing the denominator}
\\\\\\
\cfrac{4x^5}{x^{\frac{1}{3}}}+\cfrac{2x^4}{x^{\frac{1}{3}}}-\cfrac{7x^3}{x^{\frac{1}{3}}}+\cfrac{3x^2}{x^{\frac{1}{3}}}+\cfrac{9x}{x^{\frac{1}{3}}}-\cfrac{1}{x^{\frac{1}{3}}}
\\\\\\
(4x^5\cdot x^{-\frac{1}{3}})+(2x^4\cdot x^{-\frac{1}{3}})-(7x^3\cdot x^{-\frac{1}{3}})+(3x^2\cdot x^{-\frac{1}{3}})\\\\+(9x^1\cdot x^{-\frac{1}{3}})-(1\cdot x^{-\frac{1}{3}})$

$\bf 4x^{5-\frac{1}{3}}+2x^{4-\frac{1}{3}}-7x^{3-\frac{1}{3}}+9x^{1-\frac{1}{3}}-x^{-\frac{1}{3}}
\\\\\\
4x^{\frac{14}{3}}+2x^{\frac{11}{3}}-7x^{\frac{8}{3}}+9x^{\frac{2}{3}}-x^{-\frac{1}{3}}
\\\\\\
4\sqrt[3]{x^{14}}+2\sqrt[3]{x^{11}}-7\sqrt[3]{x^{8}}+9\sqrt[3]{x^{2}}-\frac{1}{\sqrt[3]{x}}$

3)

$\bf \begin{cases}
f(x)=\sqrt{x}-5x\implies &f(x)x^{\frac{1}{2}}-5x\\\\
g(x)=5x^2-2x+\sqrt[5]{x}\implies &g(x)=5x^2-2x+x^{\frac{1}{5}}
\end{cases}
\\\\\\
\textit{let's multiply the terms from f(x) by each term in g(x)}
\\\\\\
x^{\frac{1}{2}}(5x^2-2x+x^{\frac{1}{5}})\implies x^{\frac{1}{2}}5x^2-x^{\frac{1}{2}}2x+x^{\frac{1}{2}}x^{\frac{1}{5}}$

$\bf 5x^{\frac{1}{2}+2}-2x^{\frac{1}{2}+1}+x^{\frac{1}{2}+\frac{1}{5}}\implies \boxed{5x^{\frac{5}{2}}-2x^{\frac{3}{2}}+x^{\frac{7}{10}}}
\\\\\\
-5x(5x^2-2x+x^{\frac{1}{5}})\implies -5x5x^2-5x2x+5xx^{\frac{1}{5}}
\\\\\\
-25x^3+10x^2-5x^{1+\frac{1}{5}}\implies \boxed{-25x^3+10x^2-5x^{\frac{6}{5}}}$

$\bf 5\sqrt{x^5}-2\sqrt{x^3}+\sqrt[10]{x^7}-25x^3+10x^2-5\sqrt[5]{x^6}$

6 0

3 years ago

Solve for c, 3ab-2bc=12

Troyanec [42]

-2bc=12-3ab

c=(3ab-12)/(2b)

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3 years ago