Help due soon plss!!!!

Mathematics

1 answer:

Gelneren [198K]3 years ago

5 0

Answer:

Cube=150cm^2

Triangular Prism=228ft^2

Step-by-step explanation:

Cube

$A=6a^2=6*5^2=150cm^2$

Triangular prism

Using these formulas

$A=2A_{B}+(a+b+c)h\\A_{B} =\sqrt{s(s*a)(s*b)(s*c)} \\s=\frac{a+b+c}{2}$

Solving for A

$A=ah+bh+ch+\frac{1}{2} \sqrt{-a^4+2(ab)^2+2(ac)^2-b^4+2(bc)^2-c^4} \\=6.5*11+5*11+6.5*11+\frac{1}{2} *\sqrt{-6.5^4+2*(6.5*5)^2+2*(6.5*6.5)^2-5^4+2*(5*6.5)^2-6.5^4} \\=228ft^2$

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If the interval (a, infinity) describes all values of x for which the graph of f(x)=4/x^2-6x+9 is decreasing, what is the value

Afina-wow [57]

Answer:

The answer is "3".

Step-by-step explanation:

$\texttt{Given graph equation: } \\\\\bold{\Rightarrow f(x)=\frac{4}{x^2-6x+9} }$

$\bold{\ interval: (a,\infty)}$

differentiate the function f(x):

$\Rightarrow f'(x) = \frac{d }{dx}(\frac{4}{x^2-6x+9})\\\\$

Formula:

$\bold{\frac{d}{dx} \frac{v}{u}= \frac{u \frac{d}{dx} v- v\frac{d}{dx}u }{u^2}}$

$\Rightarrow f'(x) = \frac{d }{dx}(\frac{4}{x^2-6x+9})\\\\\\\Rightarrow f'(x) = \frac{d }{dx}(\frac{(x^2-6x+9) \frac{d}{dx} 4- 4\frac{d}{dx}(x^2-6x+9) }{(x^2-6x+9)^2})\\\\\\\Rightarrow f'(x) = \frac{d }{dx}(\frac{(x^2-6x+9) \times 0 - 4(2x-6) }{(x^2-6x+9)^2})\\\\\\\Rightarrow f'(x) = \frac{- 4(2x-6)}{(x^2-6x+9)^2}\\\\\\\Rightarrow \frac{- 4(2x-6)}{(x^2-6x+9)^2}=0\\\\\Rightarrow - 4(2x-6)=0\\\\\Rightarrow 8x-24=0\\\\\Rightarrow 8x=24\\\\\Rightarrow x=\frac{24}{8}\\\\\Rightarrow x=3\\\\$