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3 years ago

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Find the area and the perimeter of a rectangle with a length of 5 feet and a width of 4 5/9 feet. Write each answer as a fractio

n and as a decimal. Use bar notation if necessary. Please help me out!!

1 answer:

Over [174]3 years ago

6 0

$\bf \textit{area of a rectangle}\\\\ A=Lw~~ \begin{cases} L=5\\ w=4\frac{5}{9}\\[0.8em] \qquad \frac{4\cdot 9+5}{9}\\[0.8em] \qquad \frac{41}{5} \end{cases}\implies A=5\cdot \cfrac{41}{5}\implies \stackrel{fraction}{A=\cfrac{41}{1}}\implies \stackrel{decimal}{A=41} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \textit{perimeter of a rectangle}\\\\ A=2(L+w)\qquad \implies A=2\left(5+\cfrac{41}{5} \right) \implies A=2\left( \cfrac{25+41}{5} \right) \\\\\\ A=2\left( \cfrac{66}{5} \right)\implies A=\cfrac{132}{5}\implies \stackrel{fraction}{A=26\frac{2}{5}}\implies \stackrel{decimal}{A=26.4}$

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Maksim231197 [3]

Answer:

$C=\frac{5F-160}{9}$

Step-by-step explanation:

Let's solve for C.

$F=\frac{9C}{5}$

Step 1: Flip the equation.

$\frac{9}{5} C+32=F$

Step 2: Add $-32$ to both sides.

$\frac{9}{5} C+32+-32=F+-32$

$\frac{9}{5}C=F-32$

Step 3: Divide both sides by $\frac{9}{5}$ .

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2 years ago

Answer correctly for the brainliest answer

Dahasolnce [82]

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3 years ago

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Step-by-step explanation:

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3 years ago

Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar

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The distance between the two points is $d=8.6m$

The polar coordinate of A is $\left(4.47,296.57\right)$

The polar coordinate of B is $\left(4.24,135\right)$

Explanation:

The two points are $A(2,-4)$ and $B(-3,3)$

The distance between two points is given by,

$d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6$

Thus, the distance between the two points is $d=8.6m$

The polar coordinates of A can be written as $(Distance, tan^{-1} \frac{y}{x} )$

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Substituting $A(2,-4)$ , we get,

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To make the angle positive, let us add 360,

$\theta=360-63.43=296.57$

The polar coordinate of A is $\left(4.47,296.57\right)$

Similarly, The polar coordinate of B can be written as $(Distance, tan^{-1} \frac{y}{x} )$

Distance = $\sqrt{x^{2} +y^{2} }$

Substituting $B(-3,3)$ , we get,

Distance = $\sqrt{(3)^{2}+(3)^{2}}=4.24$

$tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45$

To make the angle positive, let us add 360,

$\theta=180-45=135^{\circ}$

The polar coordinate of B is $\left(4.24,135\right)$

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