Find the real and/or imaginary solutions by factoring

Mathematics

1 answer:

SIZIF [17.4K]2 years ago

6 0

Answer:

Step-by-step explanation:

1) (x-3),(x-8)

2) 3x^3+3x^2=27x

3 ( x 3^3+ x ^2 − 9 x − 9 )

3 ( x ^2 ( x + 1 ) − 9 ( x + 1 ) )

3 ( ( x + 1)( x ^2 − 9 ) )

3 ( ( x + 1 ) ( x ^2 − 3 ^2 ) )

3 ( x + 1 ) ( x + 3 ) ( x − 3 )

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The height of one square pyramid is 12 m. A similar pyramid has a height of 6 m. The volume of the larger pyramid is 400 m3. Det

tatiyna

$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\
$

$\bf \cfrac{smaller}{larger}\qquad \cfrac{\stackrel{height}{6}}{\stackrel{height}{12}}\implies \cfrac{1}{2}\impliedby \textit{scale factor of the pyramids}\\\\
-------------------------------\\\\
\cfrac{smaller}{larger}\qquad \cfrac{h}{h}=\cfrac{\sqrt{base}}{\sqrt{base}}\implies \cfrac{1}{2}=\sqrt{\cfrac{base}{base}}\implies \left( \cfrac{1}{2} \right)^2=\cfrac{base}{base}
\\\\\\
\cfrac{1^2}{2^2}=\cfrac{base}{base}\implies \cfrac{1}{4}=\cfrac{base}{base}\\\\
-------------------------------\\\\$

$\bf \cfrac{smaller}{larger}\qquad \cfrac{h}{h}=\cfrac{\sqrt[3]{volume}}{\sqrt[3]{volume}}\implies \cfrac{1}{2}=\sqrt[3]{\cfrac{volume}{volume}}
\\\\\\
\left( \cfrac{1}{2} \right)^3=\cfrac{volume}{volume}
\implies 
\cfrac{1^3}{2^3}=\cfrac{volume}{volume}\implies \cfrac{1}{8}=\cfrac{volume}{volume}\\\\
-------------------------------\\\\
\cfrac{smaller}{larger}\qquad \cfrac{1}{8}=\cfrac{\stackrel{volume}{v}}{\stackrel{volume}{400}}\implies \cfrac{1\cdot 400}{8}=v$

7 0

3 years ago

URGENT!!!!!!!! PLEASE HELP ASAP TIMED!!!!!!!!!!!!!!!

barxatty [35]

Step-by-step explanation:

Given that,

Perpendicular = 17 units

Hypotenuse = 25 units

We need to find the value of x.

Using Pythagoras theorem to find x.

Hypotenuse² = base²+perpendicular²

25² = x² + 17²