The equation of a line that goes through the point (x1,y1) and has a slope of m is
y-y1=m(x-x1)
so, given the point (-2,8)
x1=-2
y1=8
y-8=m(x-(-2))
y-8=m(x+2)
the equation is y-8=m(x+2) where m is the slope
or you could rewrite it to get y=mx+2m+8
Answer:
267 and 322
Step-by-step explanation:
Answer:
option B
Step-by-step explanation:
We can see in the graph that the function has two values of x where the value of y goes to infinity: x = -6 and x = 6.
These points where the value of the function goes to infinity usually are roots of the polynomial in the denominator of a fraction (when the values of x tend to these values, the denominator of the fraction tends to 0, so we have a discontinuity in the function).
So the option that represents a function that have these points in x = -6 and x = 6 is the function in option B.
The other options show functions that have only one point that goes to infinity.
Hi i a doing the same thing and i have the answer as a hope this helps
Answer:
cosjk = √55 i/3
tanjk = 8/√55 i
Step-by-step explanation:
Given
sin jk = 8/3
According to SOH CAH TOA
Sin theta = opposite/hypotenuse = 8/3
Opposite = 8
hypotenuse = 3
Get the adjacent using the pythagoras theorem
hyp² = opp²+adj²
adj² = hyp² - opp²
adj² = 3² - 8²
adj² = 9-64
adj² = -55
adj = √-55
adj = √55 i (i = √-1)
Get cosjk
cosjk = adj/hyp
cosjk = √55 i/3
Get tanjk
tanjk = opp/adj
tanjk = 8/√55 i
