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3 years ago

What are the volume formulas for the listed shapes: Cylinder, Cone and Sphere.

Mathematics

2 answers:

sesenic [268]3 years ago

8 0

Cylinder: V = $\pi r^{2} h$

Cone: V = $\pi r^{2} h /3$

Sphere: V = $4 \pi r^{3} /3$

sashaice [31]3 years ago

7 0

V of cylinder = pie × radius square × height
v of cone =1/3 × pie × radius square × height
v of a sphere = 4× pie × cubic radius all over three

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A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat

jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.

We are given a function (corrected)

$N(t) = 20(t^2-lnt^2)+ 30$

$N(t) = 20(t^2-2lnt)+ 30$

(a)

First, we take its derivative

$N'(t) = 20(2t-\frac{2}{t})$

Solve N'(t)=0

$20(2t-\frac{2}{t})=0$

Simplifying

$2t^2-2=0$

Solving for t

$t=1\ ,t=-1$

Only t=1 belongs to the valid interval $1\leqslant t\leqslant 15$

Taking the second derivative

$N''(t) = 20(2+\frac{2}{t^2})$

Which is always positive, so t=1 is a minimum

(b)

$N(1)=20(1^2-2ln1)+ 30$

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

$N(15)=20(15^2-2ln15)+ 30$

N(15)=4391.7 is a maximum

7 0

3 years ago

Help me to solve this

antoniya [11.8K]

The answer is 2cm

$8 \times 10 {}^{ - 3} \times 250 = 2000 \times 10 {}^{ - 3} \\ = 2$
or:
$8 \times {10}^{ - 3} \times 250 = 0.008 \times 250 \\ = 2$

good luck

4 0

3 years ago

The graph of a sinusoidal function has a minimum point at (0,3)(0,3) and then intersects its midline at (5π,5)

bekas [8.4K]

Answer: F(x) = 2*sin(x/10 +(3/2)*pi) + 5

Step-by-step explanation:

The information that we have is that:

We have a minimum at (0, 3)

the midline is at (5*pi, 5)

This is a sinusoidal function, so we can write one generic one as:

F(x) = A*sin(c*x + p) + B.

where A and B are constants, c is the frequency and p is a phase

First, the minimum of the sine function is when sin(x) = -1, and this happens at (3/2)*pi

We know that this minimum is at x = 0.

sin(c*0 + p) = -1

Then p = 3/2*pi.

So our function is:

F(x) = A*sin(c*x +(3/2)*pi) + B.

Now, we know that F(0) = 3, so:

3 = A*sin(c*0 +(3/2)*pi) + B = -A + B.

now we can use the other hint, the midpoint of the sine function is when sin(x) = 0, and this happens at x = 0 and x = pi, particularlly as we here have a phase of 3/2*pi, we should find x = 2*pi.

then:

c*5*pi + (3/2)*pi = 2*pi

c*5 + 3/2 = 2

c*5 = 2 - 3/2 = 1/2

C = 1/2*5 = 1/10

So our function is

F(x) = A*sin(x/10 +(3/2)*pi) + B

and we know that when x = 5*pi, F(5*pi) = 5, so:

5 = F(x) = A*sin(5*pi/10 +(3/2)*pi) + B

5 = B

and we aready knew that:

- A + B = 3

-A + 5 = 3

A = 5 - 3 = 2

So our equation is:

F(x) = 2*sin(x/10 +(3/2)*pi) + 5

8 0

3 years ago

Determine the range of the following graph:

Anni [7]

Answer:

Range is between -7 and 8

-7<y<8

Step-by-step explanation:

Range means series of number between lowest and highest output(y-axis) of a function.

Notice on the graph, the lowest y you can get is -7 and highest is 8. Other numbers are in between.

Thus, the range is in between of -7 and 8.

4 0

3 years ago

Solve -6x+3y=9, -8x+4y=12 by elimination

Lerok [7]

The answer is 678.09 plus 804

4 0

3 years ago