All categories

2 years ago

Please help me out please

Mathematics

2 answers:

grandymaker [24]2 years ago

8 0

Answer:

$\frac{500\pi }{3}$ ft³

Step-by-step explanation:

The surface area of a sphere = 4πr², hence

4πr² = 100π ( divide both sides by 4π )

r² = 25 ( take the square root of both sides )

r = $\sqrt{25}$ = 5, hence

V = $\frac{4}{3}$ πr³

= $\frac{4}{3}$ π × 5³

= $\frac{4}{3}$ π × 125 = $\frac{500\pi }{3}$ ft³

____ [38]2 years ago

6 0

Answer:

(500π)÷3

Step-by-step explanation:

area=4πr^2 , volume=4(πr^3)÷3

area=100π ,so r=5

You might be interested in

Consider the function f(x)=8-3x^2 find the value of f(x) when x= 3

scoray [572]

You start with f(x)=8-3(3)². Then, going in order of PEMDAS, you start with your exponent. So you have f(x)=8-3(9). Then multiplication, f(x)=8-27. Then your subtraction. f(x)=-19

5 0

2 years ago

Read 2 more answers

Willian bought 11 books. He decided to present 2 books to each of his friends and still have one book left. Find the number of f

ira [324]

Answer:

i think he has 5 friends

Step-by-step explanation:

sorry if im wrong

7 0

3 years ago

Read 2 more answers

How much water should be added to 3 gallons of pure acid in order to obtain the 15% acid solution? HELP DUE BY TOMMOROW MORNING

Pani-rosa [81]

3 gallons of acid = 100% concentration acid
? (water+acid) gallons = 15 concentration acid

Note: Every gallon of water added reduces the percentage concentration of acid. Therefore, the relationship is inverse proportionality.

Solution:

Total number of gallons (water+acid) = (3*100)/15 = 20 gallons
Therefore, gallons of water = 20-3 = 17 gallons

5 0

3 years ago

Ill give brainliest please help :(

OleMash [197]

Answer:

okay so god,dog,now,won,noon,doom,mood,not,ton,pool

Step-by-step explanation:

7 0

2 years ago

Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive

VikaD [51]

Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}$

recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

now, let's rationalize the denominator on tangent and secant,

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

3 0

3 years ago