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miss Akunina [59]

2 years ago

10

Someone please help me with this

1 answer:

mihalych1998 [28]2 years ago

4 0

$\frac{1}{3}\sqrt[3]{81}=\\\frac{1}{3}\sqrt[3]{{3}^{3}\times3}=\\\frac{1}{3}\sqrt[3]{{3}^{3}}\sqrt[3]{3}=\\\frac{1}{3}\times3\sqrt[3]{3}=\\\frac{3}{3}\sqrt[3]{3}=\sqrt[3]{3}$

The answer is

$\sqrt[3]{3}$

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Elena L [17]

Answer:

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Step-by-step explanation:

All we need to do is count the units in between the line segments.

3 0

1 year ago

A farmer has 520 feet of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one s

Setler [38]

Answer:

$310\text{ feet and }210\text{ feet}$

Step-by-step explanation:

GIVEN: A farmer has $520 \text{ feet}$ of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one side of the pen. The length of the barn is $310 \text{ feet}$ .

TO FIND: Determine the dimensions of the rectangle of maximum area that can be enclosed under these conditions.

SOLUTION:

Let the length of rectangle be $x$ and $y$

perimeter of rectangular pen $=2(x+y)=520\text{ feet}$

$x+y=260$

$y=260-x$

area of rectangular pen $=\text{length}\times\text{width}$

$=xy$

putting value of $y$

$=x(260-x)$

$=260x-x^2$

to maximize $\frac{d \text{(area)}}{dx}=0$

$260-2x=0$

$x=130\text{ feet}$

$y=390\text{ feet}$

but the dimensions must be lesser or equal to than that of barn.

therefore maximum length rectangular pen $=310\text{ feet}$

width of rectangular pen $=210\text{ feet}$

Maximum area of rectangular pen $=310\times210=65100\text{ feet}^2$

Hence maximum area of rectangular pen is $65100\text{ feet}^2$ and dimensions are $310\text{ feet and }210\text{ feet}$

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2 years ago

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valentinak56 [21]

Entry ticket = $7

Gift shop money = $g

Maximum to spend = $30

Inequality => 7+g ≤ 30

Hope it helps!

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1 year ago

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Mrac [35]

Answer:

what's the question???

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FrozenT [24]

x - 14 would be the equation

Does this help? Not really sure what you are asking.

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