If we are to write this equation in slope-intercept form, it will be in y = mx + b, where m is the slope of the line and b is the y intercept. We need then to find the slope of the line using 2 points on the line and filling in the slope formula to find the slope. One of the points we can use is (0, 3) which is also the y intercept. The y-intercept is found where x = 0. Where x = 0, y = 3. So b = 3. Now for the slope we will use (0,3) and (4,4): 
. Using that m value and that b value we have the equation 
. There you go!
There is 3/8 of an apple pie left from dinner. Tomorrow, Victor plans to eat 5/8 of the pie that was left. Enter how much of the
1 answer:
Answer:
of the pie
Step-by-step explanation:
Whole numbers refer to a collection of natural numbers together with 0.
Fraction refers to part of a whole.
Part of an apple pie left from a dinner =
Part of an apple eaten by Victor tomorrow is equal to of the pie that was left
So,
Part of an apple eaten by Victor tomorrow = of the pie
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Answer:
they are adjacent angles
Step-by-step explanation:
for adjacent angles
Now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
recall that
therefore, let's just plug that on the remaining ones,
now, let's rationalize the denominator on tangent and secant,
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
recall that
therefore, let's just plug that on the remaining ones,
now, let's rationalize the denominator on tangent and secant,