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Flura [38]
2 years ago
3

Which is the best estimate for the average rate of change for the quadratic function graph on the interval 4 ≤ x ≤ 12? A) 1 B) 2

C) 3 D) 4
Mathematics
1 answer:
Nonamiya [84]2 years ago
6 0

Answer:

C

Step-by-step explanation:

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You have an ant farm with 31 ants. The population of ants in your farm will double every 3 months. The table shows the populatio
Evgesh-ka [11]

Answer:

I think the table is linear. The total amount after one year is 248 ants.

Step-by-step explanation:

8 0
1 year ago
What’s the x value of PR=9x-31 and QR=43?
True [87]
<h3>Simplifying </h3>

9x + -31 = 43

<h3>Reorder the terms:</h3>

-31 + 9x =43

<h3>Solving</h3>

-31 + 9x =43

<h3>Solving for variable 'x' .</h3><h3>Move all terms containing x to the left, </h3>

Add '31' to each side of the equation .

0 + 31 +9x = 43 + 31

<h3>Combine like terms: </h3>

-31 + 31 =0

0 + 9x = 43 +31

9x = 43 + 31

<h3>Combine like terms:</h3>

43 + 31 =74

9x = 74

<h3>Divide each side by '9'</h3>

x = 8. 222222222

<h3>Simpifying</h3>

x = 8.222222222

6 0
2 years ago
Read 2 more answers
Simplify!!! <br><br>I need help please ​
satela [25.4K]

Answer:

\frac{4^{21}}{5^{6}}

Step-by-step explanation:

\frac{4^{7*3}}{5^{2*3}}=\frac{4^{21}}{5^{6}}

3 0
2 years ago
Jonathons piggy bank contains 20 nickels 30 quarters and 50 one dollar coins. He picks 20 coins from the bank at random
fredd [130]

Answer:

All in all, Jonathan's piggy bank contains 100 coins. Among these coins, only 50 are one-dollar coins. Therefore, the theoretical probability of picking one-dollar coin from the piggy bank is equal to 50/100 or 1/2.  

Similarly, from the experiment, 20 coins were picked and among these there are 12 one-dollar coins. The answer to the second question is therefore 12/20 or 3/5.

Step-by-step explanation:

4 0
2 years ago
There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. T
lesantik [10]

Answer:

150 students take physics.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the students that take calculus.

-The set B represents the students that take physics

-The set C represents the students that take chemistry.

-The set D represents the students that do not take any of them.

We have that:

A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)

In which a is the number of students that take only calculus, A \cap B is the number of students that take both calculus and physics, A \cap C is the number of students that take both calculus and chemistry and A \cap B \cap C is the number of students that take calculus, physics and chemistry.

By the same logic, we have:

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)

This diagram has the following subsets:

a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C), D

There are 360 people in my school. This means that:

a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360

The problem states that:

15 take calculus, physics, and chemistry, so:

A \cap B \cap C = 15

15 don't take any of them, so:

D = 15

75 take both calculus and chemistry, so:

A \cap C = 75

75 take both physics and chemistry, so:

B \cap C = 75

30 take both physics and calculus, so:

A \cap B = 30

Solution:

The problem states that 180 take calculus. So

a + (A \cap B) + (A \cap C) + (A \cap B \cap C) = 180

a + 30 + 75 + 15 = 180

a = 180 - 120

a = 60

Twice as many students take chemistry as take physics:

It means that: C = 2B

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

B = b + 75 + 30 + 15

B = b + 120

-------------------------------

C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)

C = c + 75 + 75 + 15

C = c + 165

----------------------------------

Our interest is the number of student that take physics. We have to find B. For this we need to find b. We can write c as a function o b, and then replacing it in the equations that sums all the subsets.

C = 2B

c + 165 = 2(b+120)

c = 2b + 240 - 165

c = 2b + 75

The equation that sums all the subsets is:

a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360

60 + b + 2b + 75 + 30 + 75 + 15 + 15 = 360

3b + 270 = 360

3b = 90

b = \frac{90}{3}

b = 30

30 students take only physics.

The number of student that take physics is:

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

B = b + 75 + 30 + 15

B = 30 + 120

B = 150

150 students take physics.

6 0
2 years ago
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