To answer the given above, multiply the whole equation with least common multiple of the denominators which is 5n. Shown below,
(4/5n - 1/5 = 2/5n) x 5n
4 - 1(n) = 2
-n = -2, n = 2
The answer is not among the choices, it could have been letter A except that the constant in A is negative.
All right...before we begin, let's lay some ground rules about the postulate your teacher asks for in every problem. <em>If you already know what they are, skip this paragraph...</em>
A postulate is a statement that claims how the triangles are congruent from looking at the paper. For example, on number 2 we have three sides of the triangle so we would use the postulate SSS (side, side, side). However, number one we would use SAS (side, angle, side) because we see the angle square showing it is a 90-degree angle.
Let's begin! (I haven't done triangles in a while so I apologize in advance if some of my answers might be incorrect on the congruent (yes or no) part of it.)
1.
A) From a visual perspective, we can see the triangles are congruent.
B) ABE = CDE because they are the corresponding points.
C) We can use SAS for the postulate.
2.
A) Yes, they are congruent.
B) OLE = OVE
C) We can use SSS for their postulate.
3.
A) Yes, they are congruent.
B) AWT = ERT
C) We can use SAS for the postulate.
4.
A) I believe the triangles are congruent. You might want to check me on that.
B) GFE = FGH
C) SSS
5.
A) They are congruent because if IH bisects it, it is directly in the middle. So, we know that WH = HS and IH = IH (duh.) and their angles match.
B) WHI = SHI
C) SAS
6.
A) This one is intriguing because it would state above the shape "LE bisects LGUE." I'm going to take that it isn't exactly in the middle, but I am still going to say it is congruent.
B) LGE = EUL
C) ASA
7.
A) Yes, they are definitely congruent.
B) RTU = TRS
C) SSS (We have nothing that indicates angles.)
8.
A) Yes, they are congruent.
B) YWV = VZY
C) We can use SAS. (Might want to check that one.)
9.
A) I would say this one is NOT congruent.
B) There are only two points. One way is HT = MA
C) They are not congruent, you can not use a postulate. However, if you teacher insists on putting one, I would use SAS.
Once again, I would be careful about my answers. I haven't worked with triangles in a few years. If my math is incorrect, or I didn't give the answer you were looking for, please let me know. However, if my math is on point, please consider marking as <em>Brainliest</em>.
Have a good one.
God Bless.
The surface area of a cylinder is the sum of the top + bottom + body, right?
The top and the bottom are circles with the same radius 3.5 . And we know that the area of a circle is r^2pi, (don't forget we have 2 circle, so, the answer must be multiplied by 2)
Now, the body: the body is just a rectangular with the width is the height of the cylinder and the length is the circumference of the circle.
The circumference = 2 r pi
the height is given
All you need to do is to calculate them and add them up. You need to work out to get the answer.
Yo don't need max height
yo use 'find vertex' equestion
since leading term is negative, this equation has a max
the vertex is the max
in the form
y=ax^2+bx+c
vertex=-b/2a
-4x^2+24x-29
a=-4
b=24
vertex=-24/(2*-4)=-24/-8=3
max height is 3 feet
dunno wat the compete square is
I can do that fo ou though
move -29 off to side and undistriute -4
-4(x^2-6x)-29
complete the square, take 1/2 of -6 an square it andadd 0
-4(x^2-6x+9-9)-29
complete square
-4((x-3)^2-9)-29
move the 9 out by distributing the -9 to the -4
36
-4(x-3)^2+36-29
-4(x-3)^2+7
answer is 3rd option
