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Tcecarenko [31]
3 years ago
6

the graph of a quadratic function is shown on the grid. which function is the best represented by this graph?

Mathematics
1 answer:
Sindrei [870]3 years ago
4 0

Answer:

Step-by-step explanation:

It is H

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Ginny is studying a population of frogs. She determines that the population is decreasing at an average rate of 3% per year. Whe
svp [43]

Answer:

Step-by-step explanation:

This is exponential, of the form

y=a(b)^x where a is the initial amount and b is the growth/decay rate. If Ginny begins the study with 1200 frogs in the population, our a value is 1200.

If the population is decreasing 3% per year, it is decreasing 100% - 3% = 97% which, as a decimal, is .97. Our exponential function, then, is

y=1200(.97)^x

6 0
3 years ago
Pls help will give Brainlyist
kvasek [131]

Answer:

B

Step-by-step explanation:

7 0
3 years ago
Order these units from smallest to largest:
Veronika [31]
The units are ordered smallest to largest from top to bottom:
Picometer (pm) = 1 x 10⁻¹² meters
Nanometer (nm) = 1 x 10⁻⁹ meters
Micrometer (um) = 1 x 10⁻⁶ meters
Millimeter (mm) = 1 x 10⁻³ meters
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Decimeter (dm) = 1 x 10⁻¹ meters
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7 0
4 years ago
How does 0.03 compare to 0.003
kati45 [8]

0.03 compare to 0.003

While comparing decimals we write the decimals one below the other according to the place value of each digits in place value chart

ones -- point -- Tenths -- Hundredths -- Thousandths

 0          .             0              3

 0          .             0              0                      3

Now we compare each digit left to right. We have 3 in hundredths  and 0 in hundredths place. 3 is greater

So 0.03  is greater than  0.003

0.03 > 0.003

6 0
4 years ago
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For x > 3, values of the function f(x) = –(x – 3)^2(x + 2) are negative. On this same interval, which statement correctly des
Andrej [43]

For x > 3, values of the function f(x) = –(x – 3)2(x + 2) are negative. On this same interval, which statement correctly describes the values of the additive and multiplicative inverses?

4 0
3 years ago
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