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3 years ago

PLLLLLLLLLLLLLLLZZZZZZ HELP IVE BEEN STUCK ON THIS ONE

Mathematics

1 answer:

Elan Coil [88]3 years ago

7 0

Answer:

yeet device and hope it breaks.

Step-by-step explanation:

Take a picture and say it died

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Question in the image

dalvyx [7]

The exact surface area is 150.8 and the exact volume is 141.37.

To answer this I used the formula for surface are for a right cylinder which is. 2•3.14•r•h+2•3.14•r^2 (^2 means to the power of 2) r=radius and h=height. When you replace the r’s and the h’s with the radius and height it’ll look like this. 2•3.14•3•5+2•3.14•3^2. For the power of 2 (^2) we basically just do 3•3. But when you put that into a calculator it gives you 150.8.

To find the volume use the formula 3.14•r^2•h when you replace the r’s and the h’s with the height and the radius it’ll look like this. 3.14•3^2•5. When put into the calculator it gives you 141.37.

7 0

3 years ago

That was my grandson question I haven’t a clue what that mean

Eduardwww [97]

That would be the zero identity property. Hope it helps!

6 0

3 years ago

The weight of a radioactive isotope was 96 grams at the start of an experiment. After one hour, the weight of the isotope was ha

zimovet [89]

The starting weight of the radioactive isotope = 96 grams

Weight after one hour is half of the starting weight. So the weight of the radioactive isotope after 1 hour = 48 grams

After 2 hours the weight is half as compared to the weight after previous hour. So weight after 2 hours = 24 grams.

This means, after every hour the weight is being halved. The half life of radioactive isotope is one hour.

Since after every hour, the weight is being halved, the weight of the isotope can be modeled by an exponential equation.

So,

Initial Weight = W₁ = 96
Change factor = 1/2 = 0.5

The general equation of the sequence will be:

$\\ \\ W_{t}=96(0.5)^{t}$

Here t represents the number of hours. Using various values of t we can find the weight of the radioactive isotope at that time.

We can plot the sequence using the above equation. The graph is attached below.

3 0

3 years ago

Read 2 more answers

What iS the common ratio of the geometric sequence of -2,4,-8,16,-32,

FromTheMoon [43]

Answer:

Step-by-step explanation:

r=4/(-2)=-2

3 0

3 years ago

A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03% chlorine. Starting at tequals0, city water conta

Katarina [22]

Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is

C(t) = 3*10⁻⁴*x(t).

The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.

Remember, 1 h = 60 minutes. The initial value problem is

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x = - 6* 10⁻⁴*(3x - 10⁻¹) x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C or |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).

Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s

7 0

3 years ago