Question

part iii - geometric and scientific applications

1. a rectangle has a perimeter of 48 cm. the length of the rectangle is (l = 2w + 4). find (l) and (w).
2. solve for (c): (f = 1.8c + 32)
3. a right triangle has legs 5 and 12. find the hypotenuse.

Explanation:

6.

Step1: Recall perimeter formula

The perimeter formula for a rectangle is $P = 2L+2W$. Given $P = 48$ and $L = 2W + 4$, substitute into the perimeter formula.
$48=2(2W + 4)+2W$

Step2: Expand and simplify

Expand the right - hand side: $48 = 4W+8 + 2W$. Combine like terms: $48=6W + 8$.

Step3: Solve for $W$

Subtract 8 from both sides: $48−8=6W$, so $40 = 6W$. Then $W=\frac{40}{6}=\frac{20}{3}\text{ cm}$.

Step4: Solve for $L$

Substitute $W=\frac{20}{3}$ into $L = 2W + 4$. $L=2\times\frac{20}{3}+4=\frac{40}{3}+4=\frac{40 + 12}{3}=\frac{52}{3}\text{ cm}$.

Step1: Isolate the term with $C$

Start with $F = 1.8C+32$. Subtract 32 from both sides: $F - 32=1.8C$.

Step2: Solve for $C$

Divide both sides by 1.8. So $C=\frac{F - 32}{1.8}$.

Step1: Apply Pythagorean theorem

For a right - triangle with legs $a = 5$ and $b = 12$ and hypotenuse $c$, the Pythagorean theorem is $a^{2}+b^{2}=c^{2}$.
Substitute $a = 5$ and $b = 12$: $5^{2}+12^{2}=c^{2}$.

Step2: Calculate the values

$5^{2}=25$ and $12^{2}=144$. Then $25 + 144=c^{2}$, so $169=c^{2}$.

Step3: Solve for $c$

Take the square root of both sides. Since $c>0$ (length of a side), $c = 13$.

Answer:

$L=\frac{52}{3}\text{ cm}$, $W=\frac{20}{3}\text{ cm}$

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