Find the area of this triangle, round to the nearest tenth, I will give brainliest

Mathematics

2 answers:

Setler79 [48]2 years ago

4 0

Answer:

59.15 in^2

Step-by-step explanation:

Using Heron's formula

semiperimeter , s = 1/2 ( 28.8+18+12) = 29.4

Area = sqrt ( 29.4(29.4-28.8)(29.4-18)(29.4-12) ) = 59.15 in^2

den301095 [7]2 years ago

3 0

Answer:

area = 58.8 in.²

Step-by-step explanation:

This is the area of a trianlge when 2 sides and the included angle are known.

Known sides: 12 in., 18 in.

Included angle: 147°

area = (1/2)ab sin C

area = (0.5)(12 in.)(18 in.)(sin 147°)

area = 58.8 in.²

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Rationalize the numerator or denominator, and simplify:

allochka39001 [22]

$\dfrac{\sqrt[3]{x+h}-\sqrt[3]x}h\times\dfrac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}=\dfrac{(\sqrt[3]{x+h})^3-(\sqrt[3]x)^3}\cdots$
$=\dfrac{x+h-x}\cdots=\dfrac h\cdots$

The $h$ s then cancel, leaving you with the $\cdots=\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}$ term.

If it's not clear what I did above, consider the substitution $a=\sqrt[3]{x+h}$ and $b=\sqrt[3]x$ . Then

$a^3-b^3=(a-b)(a^2+ab+b^2)$
$\implies\dfrac{a-b}h=\dfrac{a-b}h\times\dfrac{a^2+ab+b^2}{a^2+ab+b^2}=\dfrac{a^3-b^3}{h(a^2+ab+b^2)}$