Answer:
The choose D (2, –1)
2x-3y=4 —> 3y=2X-4 —> y= 2/3x – 4/3
y=mx+b —> So ; m= 2/3 , b= - 4/3
y-intercept :(0, -4/3)
0=2/3x –4/3 —> 2/3x = 4/3 —> X=2
x-intercept ; (2,0)
By drawing a straight line from point 2 on the x-axis and point -4/3 on the y-axis, the points that are on the axis have been extracted, but the point (2,-1) is not on the axis . :)
I hope it helped you ^_^
It is a blank picture I can’t do anything
End behavior always involves x approaching positive and negative infinity. So we'll cross off the choice that says "x approaches 1".
The graphs shows both endpoints going down forever. So both endpoints are going to negative infinity regardless if x goes to either infinity.
<h3>Answer: Choice B</h3><h3>As x approaches −∞, f(x) approaches −∞, and as x approaches ∞, f(x) approaches −∞.</h3>
Another way to phrase this would be to say "f(x) approaches negative infinity when x goes to either positive or negative infinity"
Using properties of logarithms:
log(m+n) = log(m.n)
log(m-n) = log (m/n)
we get,
log(32x16/64)
On simplifying:
log(8)
and 8= 2^3
therefore,
log(2^3)
again using another property for exponents in logarithms we get:
3 log 2 <---- Answer
