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Mekhanik [1.2K]
3 years ago
7

A television network counted 1,771,548 viewers watching the golf tournament on its channel. There were 5,861,014 viewers watchin

g a football game on another channel.
Round each number to the nearest hundred thousand.



About how many viewers were watching the two events on television?
Mathematics
1 answer:
zhenek [66]3 years ago
4 0
7,630,000 people were watching the two events on television. 
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PLEASE HELP!!!<br> Hshehebsjahsvsvabsbbsbs
IRISSAK [1]

Answer:

i think its like 9.14 or sumthin

Step-by-step explanation:

is tht NWEA? lol

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2 years ago
Solve for X. Geometry
slava [35]

Answer:  " x = 10 " .

<u>Step-by-step explanation</u>:

Note the top "line segment" ; and the bottom "line"; have equal measurements:

The length of the top line segment is:

" (2x − 8) + 5 ."

The length of the bottom line is:  " (x + 7) " ;

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To solve for "x" ;

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(2x − 8) + 5 = (x + 7) ;

Rewrite as:

2x − 8 + 5 = x + 7 ;

Note: −8 + 5 = -3 ;

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Rewrite the equation as:

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We can subtract "x" from each side of the equation;

and add "3" to each side of the equation:

 2x − 3 − x + 3 = x + 7 − x + 3 ;

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to get:

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 x = 10 .

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Let us check our answer

(2x − 8) + 5 =? (x + 7) ;

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(2*10 − 8)  + 5 =? (10 + 7) ? ;

(20 − 8)  + 5 =? 17 ?

  12 + 5 =? 17 ?  Yes!

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Best wishes in your academic pursuits!

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3 years ago
Nedd this ASAP!!<br><br><br> THANKS!
victus00 [196]

Answer:

I think is f(n) = 3 + 3(n - 1)

Step-by-step explanation:

Because it goes up 3 in price and 1 in gallons of gas

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2 years ago
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During the first stages of an epidemic, the number of sick people increases exponentially with time. Suppose that at = 0 days th
nydimaria [60]
T = days passed

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by 0 day, or t = 0, there are 2 folks sick,

\bf \qquad \textit{Amount for Exponential Growth}&#10;\\\\&#10;A=P(1 + r)^t\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &2\\&#10;P=\textit{initial amount}\\&#10;r=rate\to r\%\to \frac{r}{100}\\&#10;t=\textit{elapsed time}\to &0\\&#10;\end{cases}&#10;\\\\\\&#10;2=P(1+r)^0\implies 2=P\cdot 1\implies 2=P\qquad \boxed{A=2(1+r)^t}

 by the third day, t = 3, there are 40 folks sick,

\bf \qquad \textit{Amount for Exponential Growth}&#10;\\\\&#10;A=P(1 + r)^t\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &40\\&#10;P=\textit{initial amount}\to &2\\&#10;r=rate\to r\%\to \frac{r}{100}\\&#10;t=\textit{elapsed time}\to &3\\&#10;\end{cases}&#10;\\\\\\&#10;40=2(1+r)^3\implies 20=(1+r)^3\implies \sqrt[3]{20}=1+r&#10;\\\\\\&#10;\sqrt[3]{20}-1=r\implies 1.7\approx r\qquad \boxed{A=2(2.7)^t}

how many folks are there sick by t = 6?   \bf \stackrel{that~many}{A=2(2.7)^6}
8 0
3 years ago
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