The base of this regular right pyramid is a square. What is its surface area?

Mathematics

1 answer:

zubka84 [21]3 years ago

5 0

Answer:

○ A) 21 ft.²

Step-by-step explanation:

${a}^{2} + 2a \sqrt{ \frac{ {a}^{2}}{4} + {h}^{2} } = S. A. \\ \\ {3}^{2} + 2(3) \sqrt{ \frac{ {3}^{2} }{4} + {2}^{2}} = 24 \\ \\ 24 ≈ 21$

I am joyous to assist you anytime.

You might be interested in

1½–¾ what is the answer to this fraction

siniylev [52]

Answer:

Step-by-step explanation:

~Convert the mixed numbers into improper fractions

~Then find the LCD (Least Common Denominator) and combine

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

GL.

6 0

2 years ago

Read 2 more answers

1/8% is what as a decimal

vlabodo [156]

Answer:

$\large\boxed{\dfrac{1}{8}\%=\dfrac{1}{800}}$

Step-by-step explanation:

$p\%=\dfrac{p}{100}\\\\\dfrac{1}{8}\%=\dfrac{\frac{1}{8}}{100}=\dfrac{1}{8}\cdot\dfrac{1}{100}=\dfrac{1}{800}\\\\\text{other method}\\\\\text{Convert the fraction to the decimal:}\\\\\dfrac{1}{8}=0.125\to\dfrac{1}{8}\%=0.125\%\\\\0.125\%=\dfrac{0.125}{100}=\dfrac{0.125\cdot1000}{100\cdot1000}=\dfrac{125}{100000}=\dfrac{125:125}{100000:125}=\dfrac{1}{800}$

3 0

3 years ago

Express the complex number in trigonometric form. -6 + 6 square root of 3 i ?

bagirrra123 [75]

$\bf \begin{array}{clclll}
-6&+&6\sqrt{3}\ i\\
\uparrow &&\uparrow \\
a&&b
\end{array}\qquad 
\begin{cases}
r=\sqrt{a^2+b^2}\\
\theta =tan^{-1}\left( \frac{b}{a} \right)
\end{cases}\qquad r[cos(\theta )+i\ sin(\theta )]\\\\
-------------------------------\\\\$

$\bf r=\sqrt{(-6)^2+(6\sqrt{3})^2}\implies r=\sqrt{36+(6^2\cdot 3)}\implies r=\sqrt{144}
\\\\\\
r=12\leftarrow 
\\\\\\
\theta =tan^{-1}\left( \frac{6\sqrt{3}}{-6} \right)\implies \theta =tan^{-1}\left( -\sqrt{3}\right)\implies \theta =
\begin{cases}
\frac{2\pi }{3}\leftarrow \\
\frac{5\pi }{3}
\end{cases}$

now, notice, there are two valid angles for such a tangent, however, if we look at the complex pair, the "a" is negative and the "b" is positive, that means, "x" is negative and "y" is positive, and that only occurs in the 2nd quadrant, so the angle is in the second quadrant, not on the fourth quadrant.

thus $\bf 12\left[cos\left( \frac{2\pi }{3} \right) +i\ sin\left( \frac{2\pi }{3} \right) \right]$