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MrRissso [65]
3 years ago
10

Donkey Kong is rolling cylindrical barrels

Mathematics
1 answer:
WARRIOR [948]3 years ago
7 0
You must half the diameter to get the radius
Then you can use
2times pie times radius to get the circumference.

2 ×3.141592653589793×8.25=
51.83627878423158
You might be interested in
What is the volume of the box
Fofino [41]

Answer:

3/32 in^3

Step-by-step explanation:

volume = length * width * height

V=3/4 * 1/2 * 1/4

V= 3/32

3 0
3 years ago
Read 2 more answers
Is a + b rational or irrational
Crank

Answer and fdaStep-by-step explanation:::

The answer depends on the numbers that are inputted for a and b.

If the numbers that are inputted for both a and b are rational, then a + b would be rational.

If the numbers that are inputted for both a and b are irrational, then a + b would be irrational.

If one of the variables is rational and the other is irrational, then a + b would be irrational.

#teamtrees #PAW (Plant And Water)

6 0
2 years ago
(2x-3)(3x+4) expand and simplify
elixir [45]

Answer:

6x² - x - 12

Step-by-step explanation:

Each term in the second factor is multiplied by each term in the first factor.

(2x - 3)(3x + 4)

= 2x(3x + 4) - 3 (3x + 4) ← distribute both parenthesis

= 6x² + 8x - 9x - 12 ← collect like terms

= 6x² - x - 12 ← in expanded form


5 0
3 years ago
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

4 0
2 years ago
There are 45 new houses being built in a neighborhood. Last month, 1/3 of them were sold. This month, 1/5 of the remaining house
Makovka662 [10]
45 new houses.....last month 1/3 were sold....
1/3(45) = 45/3 = 15 houses sold last month

this month, 1/5 of the remaining houses were sold...
remaining houses left are (45 - 15) = 30
1/5(30) = 30/5 = 6 houses sold this month

houses left : 45 - 15 - 6 = 45 - 21 = 24 houses remain <==
8 0
3 years ago
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