Find the surface area of a cube with a side length of 10 2/3 in

Mathematics

1 answer:

kobusy [5.1K]3 years ago

3 0

Answer:

682 2/3

Step-by-step explanation:

(10 2/3)^2 = 113.777777777777778

113.777777777777778 (6 sides) = 682 2/3

You might be interested in

Type the correct answer in the box. Round your answer to the nearest hundredth.

HACTEHA [7]

if you add 55% and 45% the total is 100% so 45/100 are girls. 35% of the girls scored and A so 35/45 is the probability that it is a girl chosen or 7/9. the probability is 7 out of 9.

8 0

2 years ago

What is the y-intercept of y=-3x-2

andrey2020 [161]

Answer:

the answer is 2.

Step-by-step explanation:

the basic form of this equation is y=mx+c where c is the y intercept and m is the gradient. c=2 so 2 is the y intercept

5 0

3 years ago

Read 2 more answers

if the length of segment AB is 1/3 the length of segment AC and if segment AC is 12 cm long, how long is segment BC?

jeka94

The length of BC I believe will be 8

8 0

3 years ago

Read 2 more answers

Give the trigonometric ratio as a fraction: Tan Z X 24 26 Y 10 z

yulyashka [42]

Answer:

your answer is B

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the average rate of change of the function over the given interval

sattari [20]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{4}\\
t_2=\frac{3\pi }{4}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{4} \right)-h\left( \frac{\pi }{4} \right)}{\frac{3\pi }{4}-\frac{\pi }{4}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{4} \right)}{sin\left( \frac{3\pi }{4} \right)}-\frac{cos\left( \frac{\pi }{4} \right)}{sin\left( \frac{\pi }{4} \right)}}{\frac{\pi }{2}}\implies \cfrac{-1-1}{\frac{\pi }{2}}\implies \cfrac{-2}{\frac{\pi }{2}}\implies -\cfrac{4}{\pi }\\\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{3}\\
t_2=\frac{3\pi }{2}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{2} \right)-h\left( \frac{\pi }{3} \right)}{\frac{3\pi }{2}-\frac{\pi }{3}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{2} \right)}{sin\left( \frac{3\pi }{2} \right)}-\frac{cos\left( \frac{\pi }{3} \right)}{sin\left( \frac{\pi }{3} \right)}}{\frac{9\pi -2\pi }{6}}\implies \cfrac{\frac{0}{-1}-\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}{\frac{7\pi }{6}}\implies\cfrac{-\frac{1}{\sqrt{3}}}{\frac{7\pi }{6}}\implies -\cfrac{\sqrt{3}}{3}\cdot \cfrac{6}{7\pi }
\\\\\\
-\cfrac{2\sqrt{3}}{7\pi }$

8 0

3 years ago