What is the constant term in the expression 6x3y + 8x2 + 5x + 7? (Input a numeric value only.)

Mathematics

2 answers:

Jlenok [28]3 years ago

5 0

Answer: I did this quiz and it was 9 and 3 not 7.

Step-by-step explanation:

dem82 [27]3 years ago

4 0

The constant term in the expression is 7

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Ronch [10]

Answer:

A = 66.5 in²

Step-by-step explanation:

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3 years ago

Bill went to the museum at 11:30 A.M . He stayed for 3 1/2 hours.When did he leave ?

tatuchka [14]

To start, note that an hour is 60 minutes long. A 1/2 hour, or half hour, is then 60/2=30 minutes. Therefore, when we have 11 hours and 30 minutes, we have 11 and a half hours. Adding 3 and a half to that, we get 11.5+3.5=15 (a half can also be expressed as .5, although it's not typically done that way when expressing time - it just might be easier to visualize it this way). Therefore, we are 15 hours into the day. However, we can't just stop there - we have to account for AM and PM. Therefore, we subtract 12 hours from 15. If the number is positive, we are in PM - otherwise, we're in AM. Therefore, as 15-12=3, the time is in PM. The remaining number is the time, so Bill leaves at 3 PM. If we are left with a decimal (e.g. 3.25), we would keep the 3 and multiply the 0.25 (the decimal) by 60 to figure out how many minutes we have, so 3.25 would turn into 3+0.25*60=3:15.

Feel free to ask further questions!

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3 years ago

For the graph x = 12 find the slope of a line that is perpendicular to it and the slope of a line parallel to it. Explain your a

blagie [28]

Check the picture below.

that's the line of x = 12, just a straight vertical line, notice the green line, that's parallel to it, and the red line, that's perpendicular to it.

let's pick two points for each to get their slopes, hmm say for the green one (5,2) and (5,4)

$\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{2})\qquad 
(\stackrel{x_2}{5}~,~\stackrel{y_2}{4})
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{4-2}{5-5}\implies \stackrel{und efined}{\cfrac{2}{0}}$

and for the red one hmmm (3,2) and (7,2)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{2})\qquad 
(\stackrel{x_2}{7}~,~\stackrel{y_2}{2})
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-2}{7-3}\implies \cfrac{0}{4}\implies 0$

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