Please help me with this problem

Mathematics

1 answer:

tigry1 [53]3 years ago

5 0

Answer:

A- 30,968 + 31,211

Step-by-step explanation:

Since 62,179 minus 31,211 is equal to 30968, 30968 plus 31,211 can be used to help check the answer for 62,179 minus 31,211.

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Oliga [24]

The last part answers the first part for you, just look at the y-values.

In other words:

A' (-8, 2)

B' (-4, 3)

C' (-2, 8)

D' (-10, 6)

Explanation:

When you reflect any point over the x-axis, the y-value of the ordered pair is going to change.

This makes sense especially considering that the x-axis is horizontal, so the only way you could cross is to move up or down. If you were to move left or right, you'd only be able to cross the y-axis, since it's vertical.

Now for the last part, as I mentioned above, if you are reflecting across the y-axis, the x-values of the ordered pair is going to change.

A'' (8, 2)

B'' (4, 3)

C'' (2, 8)

D'' (10, 6)

Take note that the only thing that changes for the respective value is its sign, while the number itself stays the same.

3 0

3 years ago

If you use your office computer 40 hours per week, and do not unplug the computer when it is not in use, how much carbon dioxide

dangina [55]

If you use a laptop: 9.25; a desktop with a CRT monitor: 55.51; a desktop with an LCD monitor: 46.26.

A laptop that is plugged up and turned off uses 0.001kw/hr of energy. Each kwh of energy produces, on average, 1.39 lbs of CO2. There are 24*7=168 hours in the week; subtract the 40 hour work week from this and we have 128 hours a week for 52 weeks a year:
0.001*128*52=6.656*1.39=9.25.

A desktop that is plugged up and turned off uses 0.004kw/hr of energy. A CRT monitor uses 0.002 kw/hr when turned off. This means we have:
(0.004*128*52*1.39)+(0.002*128*52*1.39)=55.51.

For a desktop and an LCD monitor, which uses 0.001 kw/hr of energy, we have:
(0.004*128*52*139)+(0.001*128*52*1.39)=46.26.

6 0

4 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
\\\\\\
\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
\\\\\\
-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
\\\\\\
\textit{the initial \underline{position} was 400ft away at 0secs}
\\\\\\
-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
\\\\\\
25=t^2\implies \boxed{5=t}$

part B) check the picture below