Given:
There are given the edge of the cube-shaped aquarium has 3 feet.
Explanation:
To find the value, we need to use the volume of the cube formula:
So,
From the formula of volume:
![V=a^3](https://tex.z-dn.net/?f=V%3Da%5E3)
Where
a represents the value of edge.
So,
Put the value of edge into the above formula:
Then,
![\begin{gathered} V=a^3 \\ V=3^3 \\ V=27cube\text{ feet} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V%3Da%5E3%20%5C%5C%20V%3D3%5E3%20%5C%5C%20V%3D27cube%5Ctext%7B%20feet%7D%20%5Cend%7Bgathered%7D)
Now,
The water has a density of 62 pounds per cube foot.
According to the question:
If the weight of 1 cubic foot of water is 62 pounds, then the weight of 27 cube feet water is:
![62\times27=1674pound](https://tex.z-dn.net/?f=62%5Ctimes27%3D1674pound)
Final answer:
Hence, the water weight of the full aquarium is 1674 pounds, and the table only susupports00 pounds. So the table cannot hold the aquarium.
And,
No, the density of water would not change.
If your looking for hourly they sold a total of 12 tacos per hour.
If your looking for daily they sold a total of 288 tacos per day.
9514 1404 393
Answer:
x = 8.2
Step-by-step explanation:
As indicated on the supplied diagram, the measure of arc GH is ...
GH = 360° -83° -115° = 162°
The inscribed angle at F is half this measure, so we have ...
(10x -1)° = 162°/2
10x = 82 . . . . . . . . add 1 and simplify
x = 8.2 . . . . . . . . . divide by 10
Given that the distance between two lines in the measurement instrument is 0.01m, the maximum error should be 0.01m/2 = 0.005 m.
If you do a good work, the real measure is 1.20m +/- 0.005m, this is between 1.195 and 1.205.
Answer:
![\angle C \cong \angle M](https://tex.z-dn.net/?f=%20%5Cangle%20C%20%5Ccong%20%5Cangle%20M%20)
![\angle G \cong \angle P](https://tex.z-dn.net/?f=%20%5Cangle%20G%20%5Ccong%20%5Cangle%20P%20)
![\angle I \cong \angle R](https://tex.z-dn.net/?f=%20%5Cangle%20I%20%5Ccong%20%5Cangle%20R%20)
![\overline{CG} \cong \overline{MP}](https://tex.z-dn.net/?f=%20%5Coverline%7BCG%7D%20%5Ccong%20%5Coverline%7BMP%7D%20)
![\overline{GI} \cong \overline{PR}](https://tex.z-dn.net/?f=%20%5Coverline%7BGI%7D%20%5Ccong%20%5Coverline%7BPR%7D%20)
![\overline{CI} \cong \overline{MR}](https://tex.z-dn.net/?f=%20%5Coverline%7BCI%7D%20%5Ccong%20%5Coverline%7BMR%7D%20)
Step-by-step explanation:
Given the congruence statement ∆CGI
∆MPR, it follows that the corresponding sides of both ∆s are equal, as well as the corresponding vertices or angles. It implies that ∆CGI and ∆MPR are of the same shape and size.
✅Thus, the 6 pairs of the corresponding congruent parts of ∆CGI and ∆MPR are:
![\angle C \cong \angle M](https://tex.z-dn.net/?f=%20%5Cangle%20C%20%5Ccong%20%5Cangle%20M%20)
![\angle G \cong \angle P](https://tex.z-dn.net/?f=%20%5Cangle%20G%20%5Ccong%20%5Cangle%20P%20)
![\angle I \cong \angle R](https://tex.z-dn.net/?f=%20%5Cangle%20I%20%5Ccong%20%5Cangle%20R%20)
![\overline{CG} \cong \overline{MP}](https://tex.z-dn.net/?f=%20%5Coverline%7BCG%7D%20%5Ccong%20%5Coverline%7BMP%7D%20)
![\overline{GI} \cong \overline{PR}](https://tex.z-dn.net/?f=%20%5Coverline%7BGI%7D%20%5Ccong%20%5Coverline%7BPR%7D%20)
![\overline{CI} \cong \overline{MR}](https://tex.z-dn.net/?f=%20%5Coverline%7BCI%7D%20%5Ccong%20%5Coverline%7BMR%7D%20)