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Yanka [14]
3 years ago
13

How do you find the apothem of a pentagonal pyramid?

Mathematics
1 answer:
Archy [21]3 years ago
7 0
<span>use the formula: Volume=1/3 x(times) the area of the base x(times) height</span>
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Find an equivalent fraction 1/8 = x/56
strojnjashka [21]
<h2><u>It is 7 because 8x7 will be 56 please make this brainiest</u></h2>

4 0
3 years ago
Read 2 more answers
20 POINTS!!!
choli [55]

Hello from MrBillDoesMath!

Answer:

x = 0, 7

Discussion:

f(g(x)) =

f ( x^2-7x) =

1/ ( x^2 - 7x)


The points NOT in the domain are those where the denominator,  x^2 - 7x = 0.

x^2 - 7x   = 0       =>       factor x from each term

x(x-7) = 0            =>        one or both terms must each 0

x = 0 or x =7


Thank you,

MrB


6 0
3 years ago
2. Kira's skateboard is 71 centimeters long. What is the length of her skateboard in millimeters? Show your work. Answer:​
Marrrta [24]
Answer is 710

Divide length value by 10
7 0
2 years ago
Read 2 more answers
The new ointment was applied to four locations, and a control was applied to the other four. How many different choices were the
lubasha [3.4K]

Answer:

there are 70 possible choices for the four locations to apply the new ointment

Step-by-step explanation:

Since we have a total of 8 locations ( 4 to the new ointment and 4 to the control) ,  each one can be chosen and since the order of the locations that are chosen for the new ointment is not relevant  , then we know that the number of choices is given by the number of combinations of 4 elements in 8

number of combinations = 8 possible locations to the first ointment * 7 possible locations to the second ( since the first one was already located) * 6 to the third * 5 locations for the fourth / number of times the same combination is repeated ( the same locations but in different positions) = 8*7*6*5 / (4 possible positions for the first ointment* 3 possible positions to the second ointment (since the first one was already located * 2 possible positions of the third * 1 possible position of the fourth)

therefore

number of combinations = 8*7*6*5/(4*3*2*1 ) = 8!/((8-4)!*4!) = 70 possible combinations

thus there are 70 possible choices for the four locations to apply the new ointment

6 0
3 years ago
Find the exact value of cos(sin^-1(-5/13))
son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0
3 years ago
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