F^3 x f^9 x f^4 Simplify

Mathematics

1 answer:

Hatshy [7]3 years ago

6 0

Answer:

f¹⁶

Step-by-step explanation:

f³*f⁹*f⁴

f⁽³⁺⁹⁺⁴⁾

f¹⁶

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Suppose you want to build a small walkway around your garden so that there is someplace to walk around and work on your vegetabl

wolverine [178]

Answer:

Approximately 3.5 feet - Option B

Step-by-step explanation:

Imagine that you have this walkway around the garden, with dimensions 30 by 20 feet. This walkway has a difference of x feet between it's length, and say the dimension 30 feet. In fact it has a difference of x along both dimensions - on either ends. Therefore, the increases length and width should be 30 + 2x, and 20 + 2x, which is with respect to an increases area of 1,000 square feet.

( 30 + 2x ) $*$ ( 20 + 2x ) = 1000 - Expand "( 30 + 2x ) $*$ ( 20 + 2x )"

600 + 100x + 4 $x^2$ = 1000 - Subtract 1000 on either side, making on side = 0

4 $x^2$ + 100x - 400 = 0 - Take the "quadratic equation formula"

( Quadratic Equation is as follows ) - $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ,

$x=\frac{-100+\sqrt{100^2-4\cdot \:4\left(-400\right)}}{2\cdot \:4}:\quad \frac{5\left(\sqrt{41}-5\right)}{2}$ ,

$x=\frac{-100-\sqrt{100^2-4\cdot \:4\left(-400\right)}}{2\cdot \:4}:\quad -\frac{5\left(5+\sqrt{41}\right)}{2}$

There can't be a negative width of the walkway, hence our solution should be ( in exact terms ) $\frac{5\left(\sqrt{41}-5\right)}{2}$ . The approximated value however is 3.5081...or approximately 3.5 feet.