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aleksklad [387]
3 years ago
11

Question 83 true or false and why ? plz ryt answer thanks

Mathematics
1 answer:
Vikki [24]3 years ago
6 0
The answer is false.

The formula for the SA of a cone is \pi r^{2}+\pi rl where <em>r</em> is the radius of the base and <em>l</em> is the slant height.
Since we have both the slant's height and the radius of the base we can answer this just by plugging numbers in;
SA=\pi r^{2}+\pi rl\\&#10;SA=\pi (2^{2})+\pi (2)(6)\\&#10;SA=4\pi+12\pi\\&#10;SA=16\pi\\
16\pi≈50.2654...

So your answer is false.
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In the picture, the angle made by the goniometer is classified as a(n) ___ angle. ___ ° &lt; θ &lt; ___ .
Svet_ta [14]
An obtuse angle which is greater than 90 degrees and less than 180 degrees.

5 0
3 years ago
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Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
3 years ago
Find an equation for the line tangent to the curve at the point defined by the given value of d²y/dx².​
mars1129 [50]

Answer:

Step-by-step explanation:

Given:

x = 2cost,

t = (1/2)arccosx

y = 2sint

dy/dx = dy/dt . dt/dx

dy/dt = 2cost

dt/dx = -1/√(1 - x²)

dy/dx = -2cost/√(1 - x²)

Differentiate again to obtain d²y/dx²

d²y/dx² = 2sint/√(1 - x²) - 2xcost/(1 - x²)^(-3/2)

At t = π/4, we have

(√2)/√(1 - x²) - (√2)x(1 - x²)^(3/2)

4 0
2 years ago
You are studying for your final exam of the semester. up to this point, you received 3 exam scores of 67%, 68%, and 78℅. to rece
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213 + x = 79 * 4
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lowest u can make is 67, highest u can make is 100
4 0
2 years ago
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Nataly [62]

Let's solve the equation:

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Infinitely many solutions would be correct because no matter what x is, it will always equal each other the both sides of the equation because it is 9 times x on both sides.


7 0
2 years ago
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