we are given

now, we can compare it with

we can find b
we get

now, we are given
How would the graph change if the b value in the equation is decreased but remains greater than 1
Let's take
b=1.8

b=1.6

b=1.4

b=1.2

now, we can draw graph
now, we will verify each options
option-A:
we know that all y-value will begin at y=0
because horizontal asymptote is y=0
so, this is FALSE
option-B:
we can see that
curve is moving upward when b decreases for negative value of x
but it is increasing slowly for negative values of x
so, this is FALSE
option-C:
we can see that
curve is moving upward when b decreases for negative value of x
but it is increasing slowly for negative values of x
so, this is TRUE
option-D:
we know that curves are increasing
so, the value of y will keep increasing as x increases
so, this is TRUE
option-E:
we can see that
curve is moving upward when b decreases for negative value of x
but it is increasing slowly for negative values of x
so, this is FALSE
This basically says all the steps ! If you have any questions just tell me :)
Well I would say the answer to your question is
A why?
well because -4 times -2 equals Positive 8 BUT if you times 8 by negative 1 it turns it back into a negative number!
Hope this helps
Answer:
289/216
Step-by-step explanation:
add the parentheses first
(2/3+3/4)×(3/6+4/9)=
17/12×17/18
then multiply
17/12×17/18=
286/216
The solution is 2sqrt77/77. You just fill in 2/3 for the x's. When you do that, you do that you get (2/3)/sqrt(9-[2/3]^2) which, simplified, is (2/3)/sqrt(9-[4/9]). Now use the common denominator under the radical of 9 to get (2/3)/sqrt([81-4]\9). Simplifying even further gives you (2/3)/([sqrt(77)]/3). Now do that division by multiplying 2/3 by the reciprocal of ([sqrt(77)]/3) to get 2/sqrt77. I rationalized the denominator to get that result up there.