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alina1380 [7]
3 years ago
6

Calculate the answer. Express It in scientific notation. All answers should have the correct number of significant figures.

Mathematics
1 answer:
ad-work [718]3 years ago
5 0

Calculate the answer. Express It in scientific notation. All answers should have the correct number of significant figures.

(4.9 x 102) (9.80 x 102) =

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Which statement best demonstrates why the following is a non-example of a polynomial?
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The following statements <span>demonstrates why the following is a non-example of a polynomial.</span>
1. The expression has a variable raised to a negative exponent. 
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3. The expression has a variable raised to a fraction. 
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Solve the equation using the quadratic formula. x^2-4x+3=0
almond37 [142]
Let's solve your equation step-by-step.

x2−4x+3=0

Step 1: Factor left side of equation.

(x−1)(x−3)=0

Step 2: Set factors equal to 0.
x-1=0 or x-3=0

For this one add one on both sides
x−1+1=0+1 x=1
For this one add 3 on both sides
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3 years ago
22. (01.05)
shusha [124]

Answer:

Step-by-step explanation:

The range would only include positive integers.

4 0
3 years ago
Greg mowed 7 yards and received the same amount of money for each yard he mowed. He made a total of $210. (If y is the number of
Rudik [331]

Answer:

7y=$210

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
In 1981, the Australian humpback whale population was 350 and has increased at a rate of 14% each year since then.)
dimulka [17.4K]

Answer:

In 1981, the Australian humpback whale population was 350

Po = Initial population = 350

rate of increase = 14% annually

P(t) = Po*(1.14)^t

P(t) = 350*(1.14)^t

Where

t = number of years that have passed since 1981

Year 2000

2000 - 1981 = 19 years

P(19) = 350*(1.14)^19

P(19) = 350*12.055

P(19) = 4219.49

P(19) ≈ 4219

Year 2018

2018 - 1981 = 37 years

P(37) = 350*(1.14)^37

P(37) = 350*127.4909

P(37) = 44621.84

P(37) ≈ 44622

There would be about  44622  humpback whales in the year 2018

5 0
4 years ago
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