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3 years ago

The RADIUS of the barrel is 4. What is the area of the barrel?

Mathematics

1 answer:

Lesechka [4]3 years ago

5 0

It should be roughly 50.27? when finding the area of a circle, be sure to use a=πr^2!

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The population of a colony of bacteria can be modeled by the function P(t) = 19,300(5)^t, where t is the elapsed time

allsm [11]

Answer: root t(5)/19300

Explanation:

P(t) = 19300(5)^t
P-1(t) = root t(5)/19300

4 0

4 years ago

Read 2 more answers

7.05, 5.07,7.5,0.57 least to greatest

Rudiy27

0.57 , 5.07 , 7.05 , 7.5

7 0

3 years ago

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Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.

Elden [556K]

Answer:

The answer is " $cirtical\ points \ (x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$ "

Step-by-step explanation:

Given:

$g(x) = (x-e)^3e^{-(x-e)}$

Find critical points:

$g(x) = (x-e)^3e^{(e-x)}$

differentiate the value with respect of x:

$\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r} \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]$

critical points $g'(x)=0$

$\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3$

So,

The critical points of $(x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$

7 0

3 years ago

#11 and #13 geometry help please help need work shown as well !!

Trava [24]

Answer: In diagram (11) X = 30 degrees and Y = 60 degrees

In diagram (13) X = 20.55 degrees and Y = 69.45 degrees

Step-by-step explanation: What we have are two right angled triangles with two sides given in each case, therefore we can calculate the two unknown angles (since the third angle equals 90 degrees).

For the second triangle (number 13) we shall use Y as the reference angle. That means we have an opposite (the side facing the reference angle) which is 24 and we have an adjacent (the side that lies between the right angle and the reference angle) which is 9. Hence,

Tan Y = opposite/adjacent

Tan Y = 24/9

Tan Y = 2.6667

Checking with your calculator or table of values, 2.6667 = 69.45

Having two angles, which are 90 and 69.45, the third angle X can be computed as

X = 180 - (90 + 69.45)

Sum of angles in a triangle equals 180

X = 180 - 159.45

X = 20.55 and Y = 69.45

For the first triangle (number 11) we shall use X as the reference angle. We have two sides, the opposite (the side facing the reference angle) which is 3, and the hypotenuse (side facing the right angle) which is 6. Hence,

Sin X = opposite/hypotenuse

Sin X = 3/6

Sin X = 0.5000

Checking with your calculator or table of values,

X = 30 degrees.

Also, now that we have two angles we can easily compute the third angle as follows

Y = 180 - (90 + 30)

Sum of angles in a triangle equals 180

Y = 180 - 120

Y = 60

X = 30 and Y = 60

4 0

3 years ago

Given the vectors A⃗ and B⃗ shown in the figure ((Figure 1) ), determine the magnitude of B⃗ −A⃗. A is 28 degrees above the posi

Vlad [161]

This problem is represented in the Figure below. So, we can find the components of each vector as follows:

$\bullet \ cos(28^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{A_{x}}{44} \\ \\ \therefore A_{x}=44cos(28^{\circ})=38.85m \\ \\ \\ \bullet \ sin(28^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{A_{y}}{44} \\ \\ \therefore A_{y}=44sin(28^{\circ})=20.65m$

$\bullet \ cos(56^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{-B_{x}}{26.5} \\ \\ \therefore B_{x}=-26.5cos(56^{\circ})=-14.81m \\ \\ \\ \bullet \ sin(56^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{B_{y}}{26.5} \\ \\ \therefore B_{y}=26.5sin(56^{\circ})=21.97m$

Therefore:

$\vec{A}=(38.85, 20.65)m \\ \\ \vec{B}=(-14.81, 21.97)m$

So:

$\vec{B}-\vec{A}=(-14.81, 21.97)-(38.85, 20.65)=(-53.66,1.32)$

Finally, the magnitude is:

$\boxed{\left| \vec{B}-\vec{A}\right|=\sqrt{(-53.66)^2+(1.32)^2}=53.67m}$

7 0

3 years ago

Read 2 more answers