Answer:Assuming all three, we shall find that each of the relations in 3:14 leads to a ... Then by 3:15 the relations AD//BC and AB||DE imply AD//CE, which excludes ... From 2:72, 3:11, 3:14, and 3:16 we deduce 3:19 If A, B, C are three distinct ... a point D lies between X and Y in AB/C if it belongs to XY/C, that is, if XY||CD
Step-by-step explanation:
Answer:
thats quadrant 2
Step-by-step explanation:
Answer:
68
Step-by-step explanation:
because 180-44=136
136:2=68 (since TRS=TSR, there is formula)
Short answer: I don't know, but that doesn't mean I can't give you something that you can decide for yourself.
y = 4*2^(2n - 2) is the pattern.
Go for broke. Try n = 4. You should get 256. Let's try it.
y = 4 * 2^(2*4 - 2)
y = 4 * 2^(8 - 2)
y = 4 * 2^6
y = 4 * 64
y = 256 yup it works.
The other end is just as important. Suppose n = 1
Then y = 4 * 2^(2*1 - 2) = 4 * 2^0 = 4*1 = 4 Both work.
If this formula is correct, we can abbreviate it to make your task easier.
y = 4 * 2^(2n - 2)
y = 2^2 * 2^(2n - 2)
y = 2^(2n - 2 + 2)
y = 2^(2n) Now try the two end points again.
n = 4
y = 2^(2*4)
y = 2^8
y = 256
n = 1
y = 2^(2*1)
y = 2^2
y = 4 which again checks.
so y = 2^(2n) I think is an exponential function.
Sorry my explanation is so long.
Answer:
a =
Step-by-step explanation:
Given:
f(x) = log(x)
and,
f(kaa) = kf(a)
now applying the given function, we get
⇒ log(kaa) = k × log(a)
or
⇒ log(ka²) = k × log(a)
Now, we know the property of the log function that
log(AB) = log(A) + log(B)
and,
log(Aᵇ) = b × log(A)
Thus,
⇒ log(k) + log(a²) = k × log(a) (using log(AB) = log(A) + log(B) )
or
⇒ log(k) + 2log(a) = k × log(a) (using log(Aᵇ) = b × log(A) )
or
⇒ k × log(a) - 2log(a) = log(k)
or
⇒ log(a) × (k - 2) = log(k)
or
⇒ log(a) = (k - 2)⁻¹ × log(k)
or
⇒ log(a) =
(using log(Aᵇ) = b × log(A) )
taking anti-log both sides
⇒ a =