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otez555 [7]
3 years ago
7

HELP WITH GEO!!! PLS 50 POINTS

Mathematics
2 answers:
pashok25 [27]3 years ago
5 0

Answer:

1. A   2. B    3. A

Step-by-step explanation:

photoshop1234 [79]3 years ago
3 0

Answer:

1.A

2.B

3.A

I believe these would be the correct answers

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A box contains 6 red, 3 white, 2 green, and 1 black (total 12) identical balls. What is the least number of balls nevessary to t
anzhelika [568]

Answer:

7 balls

Step-by-step explanation:

Look for the worst case scenario:

Get 3 white, 2 green, and 1 black.

That adds up to: 3 + 2 + 1 = 6.

+1 red ball: 6 + 1 =7

7 balls

5 0
2 years ago
Factor out the GCF of 16x3−4x2+24x
Anni [7]
The GCF is 4x

4x (4x^2 - x + 6)
6 0
3 years ago
What is the value of b
n200080 [17]

Angle = 68⁰ is an inscribed angles, this angle equal 68⁰,

so it intersect arc = 2*68⁰= 136⁰.

Arc of the whole circle (circumference) =360⁰,and it also =b+104+136.

b+104+136 =360

b=360-(104+136) = 120⁰

Answer A. b=120⁰.

7 0
3 years ago
Line m is drawn so that it is perpendicular to two distinct planes, Q and R. What must be true about planes Q and R?
UNO [17]
B. are parallel.

<span>Line m is drawn so that it is perpendicular to two distinct planes, Q and R. Therefore, planes Q and R are parallel. 

Since it doesn't state that two planes intersect with each other, it is safe to assume that only line m is their connection. The planes are parallel with each other. </span>
7 0
3 years ago
Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li
krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years

First, let us calculate the decay constant (k)

75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036

Next, let us calculate the half-life as follows:

\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2}  \approx 669\ years

Therefore the half-life is between 600 and 700 years

5 0
3 years ago
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