First - 8
second - 4
third - 12
fourth - 9
Answer:
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Step-by-step explanation:
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Answer:
15/22
Step-by-step explanation:
Answer:
![A=189\ mm^2](https://tex.z-dn.net/?f=A%3D189%5C%20mm%5E2)
Step-by-step explanation:
<u>Surface Areas
</u>
Is the sum of all the lateral areas of a given solid. We need to compute the total surface area of the given prism. It has 5 sides, two of them are equal (top and bottom areas) and the rest are rectangles.
Computing the top and bottom areas. They form a right triangle whose legs are 4.5 mm and 6 mm. The area of both triangles is
![\displaystyle A_t=2*\frac{b.h}{2}=b.h=(4.5)(6)=27 mm^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A_t%3D2%2A%5Cfrac%7Bb.h%7D%7B2%7D%3Db.h%3D%284.5%29%286%29%3D27%20mm%5E2)
The front area is a rectangle of dimensions 7.7 mm and 9 mm, thus
![A_f=b.h=(7.5)(9)=67.5 \ mm^2](https://tex.z-dn.net/?f=A_f%3Db.h%3D%287.5%29%289%29%3D67.5%20%5C%20mm%5E2)
The back left area is another rectangle of 4.5 mm by 9 mm
![A_l=b.h=(4.5)(9)=40.5 \ mm^2](https://tex.z-dn.net/?f=A_l%3Db.h%3D%284.5%29%289%29%3D40.5%20%20%5C%20mm%5E2)
Finally, the back right area is a rectangle of 6 mm by 9 mm
![A_r=b.h=(6)(9)=54 \ mm^2](https://tex.z-dn.net/?f=A_r%3Db.h%3D%286%29%289%29%3D54%20%5C%20mm%5E2)
Thus, the total surface area of the prism is
![A=A_t+A_f+A_l+A_r=27+67.5+40.5+54=189\ mm^2](https://tex.z-dn.net/?f=A%3DA_t%2BA_f%2BA_l%2BA_r%3D27%2B67.5%2B40.5%2B54%3D189%5C%20mm%5E2)
![\boxed{A=189\ mm^2}](https://tex.z-dn.net/?f=%5Cboxed%7BA%3D189%5C%20mm%5E2%7D)
Answer:
![y=-\frac{1}{5} x-5](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B5%7D%20x-5)
Step-by-step explanation:
The given line is defined by:
, where we see that the slope is 5 and the y-intercept 1.
In order to find a line perpendicular to the given one, we need it to have a slope that is the "opposite of the reciprocal" of the given slope.
"Opposite" means it would have its sign inverted (in our case from positive to negative); and "reciprocal means that instead of 5, it would be its reciprocal:
.
We can write this new line with such slope, and try to find its y-intercept (b) by using the given condition that requires it to go through the point (-5,-4) on he plane:
![y=-\frac{1}{5} x+b](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B5%7D%20x%2Bb)
we require then that when
, the value of
.
Therefore: ![-4=-\frac{1}{5} (-5)+b\\-4=\frac{5}{5} +b\\-4=1+b\\b=-4-1=-5](https://tex.z-dn.net/?f=-4%3D-%5Cfrac%7B1%7D%7B5%7D%20%28-5%29%2Bb%5C%5C-4%3D%5Cfrac%7B5%7D%7B5%7D%20%2Bb%5C%5C-4%3D1%2Bb%5C%5Cb%3D-4-1%3D-5)
Then our final answer is that the new line should have the form: ![y=-\frac{1}{5} x-5](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B5%7D%20x-5)