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Brrunno [24]
3 years ago
10

In 2013, the population of the state of New York was approximately 19.65 million, and the population of New York City was 8,406,

000. In 2013, how many people in New York state did not live in New York City?
The answer in standard notation is BLANK. The answer in scientific notation is p × 10q, where p is BLANK and q is BLANK

PLS FILL IN THE BLANKS
Mathematics
2 answers:
Anika [276]3 years ago
4 0
For the first blank you would get an answer of 11,244,000
The second blank would be 11.244 x 10^6

I found the first answer by subtracting 8,406,000 from 19,650,000

The second answer by converting the first answer into a decimal and counting how many decimal places I would have to move in order to get the original number
steposvetlana [31]3 years ago
3 0

Answer:

huhhhhhhhhhh

Step-by-step explanation:

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Which is larger 1 1/2 or 1 15/16
solniwko [45]

Hi there!  

»»————- ★ ————-««

I believe your answer is:  

1 (15/16)

»»————- ★ ————-««  

Here’s why:  

⸻⸻⸻⸻

\boxed{\text{Converting the Fractions...}}\\\\\rightarrow \frac{1}{2}=\frac{1*8}{2*8}=  \boxed{\frac{8}{16}}\\\\\text{We would only compare the fractions because we have the same whole number.}\\\\\frac{8}{16} 1\frac{1}{2}}

⸻⸻⸻⸻

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.  

6 0
2 years ago
Read 2 more answers
Does anyone know how to do this help please
statuscvo [17]

Answer:

\boxed{\begin{array}{c|c|c|c|c|c|c|c} \bf x &\rm -3 &\rm -2 &\rm -1 &\rm 0 &\rm 1 &\rm 2 &\rm 3 \\\\ \bf y & \rm 0 &\rm -4 &\rm -6&\rm -6 &\rm -4&\rm 2 &\rm 6 \end{array}}

Step-by-step explanation:

A quadratic function is given to us . And we need to fill out the table by using the function . The given function is ,

\rm\implies y = x^2+x - 6

Here we need to substitute the different values of x , to get the different values of y.

<u>Put</u><u> </u><u>x </u><u>=</u><u> </u><u>-</u><u>3</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 3^2-3-6\\

\rm\implies y = 9 - 3 - 6 \\

\rm\implies y = 0

<u>Put </u><u>x </u><u>=</u><u> </u><u>-</u><u>2</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 2^2-2-6\\

\rm\implies y = 4 -2-6\\

\rm\implies y = -4

<u>Put</u><u> </u><u>x </u><u>=</u><u> </u><u>-</u><u>1</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = -1^2-1-6\\

\rm\implies y = 1 -1-6 \\

\rm\implies y = -6

<u>Put </u><u>x </u><u>=</u><u> </u><u>1</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 1^2+1-6\\

\rm\implies y = 2 -6

\rm\implies y = -4

<u>Put </u><u>x </u><u>=</u><u> </u><u>2</u><u> </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 2^2+2-6\\

\rm\implies y = 4 +2-6\\

\rm\implies y = 2

<u>Put </u><u>x </u><u>=</u><u> </u><u>3 </u><u>:</u><u>-</u><u> </u>

\rm\implies y = 3^2+3-6\\

\rm\implies y = 9 +3-6\\

\rm\implies y = 6

<u>Final </u><u>table</u><u> </u><u>:</u><u>-</u><u> </u>

\boxed{\begin{array}{c|c|c|c|c|c|c|c} \bf x &\rm -3 &\rm -2 &\rm -1 &\rm 0 &\rm 1 &\rm 2 &\rm 3 \\\\ \bf y & \rm 0 &\rm -4 &\rm -6&\rm -6 &\rm -4&\rm 2 &\rm 6 \end{array}}

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