How do you find the square root of a decimal for example 0.36
1 answer:
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when running a line, in a right-triangle, from the 90° angle perpendicular to its opposite side, we will end up with three similar triangles, one Small, one Medium and a containing Large one. Check the picture below.
solution: option B and C both are correct i.e., option C is correct i.e., ∠E ≅∠H and ∠I ≅ ∠F .
option C is correct i.e., ∠E ≅∠H.
explanation:
it is given that ratio of corresponding sides of ΔFGE and ΔIJH are equal
i.e.,
and if ∠E ≅ ∠H
Then ΔFGE and ΔIJH are similar by SAS (side angle side) similarity theorem.
so option C is correct i.e., ∠E ≅ ∠H.
and option B is also correct
explanation:
since it is given that
And if ∠I ≅ ∠F
then ΔFGE and ΔIJH are similar by SAS (side angle side) similarity theorem.
Zeroes
set numerator to zero
x^2=0
x=0
zeroes are at (0,0)
set numerator to zero
x^2=0
x=0
zeroes are at (0,0)
HA
for p(x)/q(x)
when degree of p(x)<q(x), HA=0
when degree of p(x)=q(x), HA= leading coef of p(x) divided by leading coef of q(x)
when degree of p(x)>q(x) you probably have a slant assymtote
degrees are same (no slant assymtote)
leading coefs
1/1=1
HA is y=1
does it cross?
1=(x^2)/(x^2-4)
x^2-4=x^2
minus x^2 both sides
-4=0
false, it does not cross the HA
VA
simplify the fraction (it can't)
set the denomenator to zero
(x-2)(x+2)=0
VA's at x=2 and x=-2
so to graph
graph the point (0,0)
draw the lines
y=1
x=-2
x=2
reemmber, plus, minus, plus
so from left, it goes from above the HA right up to the VA of x=-2
then goes upside down U shape in between VA's going through (0,0)
then from top of VA x=2 down to y=1
then gets closer to y=1 but never touches