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3 years ago

Geometry help please

Mathematics

1 answer:

Nikolay [14]3 years ago

3 0

x
------- = 19.32
50

x
(50) ----- = 19.32 (50)
50

x = 966 cm^3

Hope this helps! :)

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You must design a closed rectangular box of width w, length l and height h, whose volume is 504 cm3. The sides of the box cost 3

Charra [1.4K]

Answer:

Dimensions will be

Length = 7.23 cm

Width = 7.23 cm

Height = 9.64 cm

Step-by-step explanation:

A closed box has length = l cm

width of the box = w cm

height of the box = h cm

Volume of the rectangular box = lwh

504 = lwh

$h=\frac{504}{lw}$

Sides which involve length and width and height, cost = 3 cents per cm²

Top and bottom of the box costs = 4 cents per cm²

Cost of the sides $C_{s}$ = 3[2(l + w)h] = 6(l + w)h

$C_{s}$ = 3[2(l + w)h]

$C_{s}=6(l+w)(\frac{504}{lw} )$

Cost of the top and the bottom $C_{(t,p)}$ = 4(2lw) = 8lw

Total cost of the box C = $3024\frac{(l+w)}{lw}$ + 8lw

= $3024[\frac{1}{l}+\frac{1}{w}]$ + 8lw

To minimize the cost of the sides

$\frac{dC}{dl}=3024(-l^{-2}+0)+8w=0$

$\frac{3024}{l^{2}}=8w$

$\frac{378}{l^{2}}=w$ ---------(1)

$\frac{dC}{dw}=3024(-w^{-2})+8l=0$

$\frac{3024}{w^{2}}=8l$

$\frac{378}{w^{2}}=l$

$w^{2}=\frac{378}{l}$ -------(2)

Now place the value of w from equation (1) to equation (2)

$(\frac{378}{l^{2}})^{2}=\frac{378}{l}$

$\frac{(378)^{2} }{l^{4}}=\frac{378}{l}$

l³ = 378

l = ∛378 = 7.23 cm

From equation (2)

$w^{2}=\frac{378}{7.23}$

$w^{2}=52.28$

w = 7.23 cm

As lwh = 504 cm³

(7.23)²h = 504

$h=\frac{504}{(7.23)^{2}}$

h = 9.64 cm

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Answer:

a) x^2 + y^2

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Step-by-step explanation:

Here, we want to write the algebraic statements as expressions;

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The square of y is y^2

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Answer:

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Step-by-step explanation:

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$\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x
&& y={{ k }}x
\end{array}\\ \quad \\
$

and also

$\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
&&y=\cfrac{{{ k}}}{x}
\end{array}
$

now, we know that V varies directly to T and inversely to P simultaneously
thus $\bf V=T\cdot \cfrac{k}{P}$

so $\bf V=T\cdot \cfrac{k}{P}\qquad 
\begin{cases}
V=42\\
T=84\\
P=8
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\\\\\\
V=\cfrac{4T}{P}\qquad now\quad 
\begin{cases}
V=74\\
P=10
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