Answer:
Yes
Step-by-step explanation:
The HL theorem basically says that two triangles are congruent if their corresponding hypotenuses and one leg are equal.
Here, the hypotenuse of both triangles are each marked with two dashes meaning they're equal. Similarly, QE and ET are both equal.
Therefore we can see that they can be proved congruent by the HL Theorem.
X = 0.4/3
x = 13
if answer is wanted in decimal form.
![\bf \cfrac{(x-2)(x+3)}{2x+2}\implies \cfrac{x^2+x-6}{2x+2}~~ \begin{array}{llll} \leftarrow \textit{2nd degree polynomial}\\ \leftarrow \textit{1st degree polynomial} \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{vertical asymptote}}{2x+2=0}\implies 2x=-2\implies x=-\cfrac{2}{2}\implies x=-1](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%28x-2%29%28x%2B3%29%7D%7B2x%2B2%7D%5Cimplies%20%5Ccfrac%7Bx%5E2%2Bx-6%7D%7B2x%2B2%7D~~%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cleftarrow%20%5Ctextit%7B2nd%20degree%20polynomial%7D%5C%5C%20%5Cleftarrow%20%5Ctextit%7B1st%20degree%20polynomial%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bvertical%20asymptote%7D%7D%7B2x%2B2%3D0%7D%5Cimplies%202x%3D-2%5Cimplies%20x%3D-%5Ccfrac%7B2%7D%7B2%7D%5Cimplies%20x%3D-1)
when the degree of the numerator is greater than the denominator's, then it has no horizontal asymptotes.
quick note:
when the degree of the numerator is 1 higher than the degree of the denominator, then it has an slant-asymptote, so this one has a slant-asymptote.
Answer:
The volume of the cylinder is 25.12
Step-by-step explanation:
The width is the diameter in the circle of the base
radius = half of diameter
r = 2/2 = 1
To calculate the volume of a cylinder we have to use the following formula:
v = volume
h = height = 8
π = 3.14
r = radius = 1
v = (π * r²) * h
we replace the unknowns with the values we know
v = (3.14 * (1)²) * 8
v = (3.14 * 1) * 8
v =3.14 * 8
v = 25.12
The volume of the cylinder is 25.12