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3 years ago

I need help 1. 2. 3.

Mathematics

1 answer:

Reptile [31]3 years ago

4 0

1. North America, Africa, Asia, Europe

2. South America, Antarctica, and Australia

3. North America, South America and Antarctica.

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Jose's football weighs 7/8 pound. His football helmet weighs 5 1/6 pounds. Estimate how much more the helmet weighs than the foo

PtichkaEL [24]

Answer:

4 whole number, 7/24

Step-by-step explanation:

Thats correct i didnt had any explination because i wrote it on a paper

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Step-by-step explanation:

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3 years ago

Please help me find the area

photoshop1234 [79]

Answer:

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Step-by-step explanation:

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stich3 [128]

<u>The answer is -84.</u>

Steps:

1. Follow PEMDAS as seen in order of operations.

2: Calculate within parenthesis.

3: Multiply & Divide (Left to right.)

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3 years ago

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The population of California was 29.76 million in 1990 and 33.87 million in 2000. Use the model found in part a (f(x) =29.76(1.0

NeTakaya

Answer:

The average rate of change from 1990 to 1991 is $0.387 \:\frac{million}{years}$

The average rate of change from 2000 to 2001 is $0.440 \:\frac{million}{years}$

Step-by-step explanation:

The average rate of change of function f(x) over the interval $a\leq x\leq b$ is given by

$\frac{f(b)-f(a)}{b-a}$

It is a measure of how much the function changed per unit, on average, over that interval.

From the information given:

The function that models the population t years after 1990 is $f(t) =29.76(1.013)^t$
The year 1990 is t = 0 and the year 2000 is t = 10.

1. The average rate of change from 1990 to 1991 is:

The interval is $0\leq x\leq 1$

$\frac{29.76(1.013)^1-29.76(1.013)^0}{1-0}\\\\\frac{1.013^1\cdot \:29.76-1\cdot \:29.76}{1-0}\\\\\frac{0.38688}{1-0}\\\\0.387 \:\frac{million}{years}$

2. The average rate of change from 2000 to 2001 is

The interval is $10\leq x\leq 11$

$\frac{29.76(1.013)^{11}-29.76(1.013)^{10}}{11-10}\\\\\frac{0.44022}{11-10}\\\\0.440 \:\frac{million}{years}$

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3 years ago