Answer:
f(x) approaches positive infinity.
g(x) approaches negative infinity.
the y-intercept of f(x) is less than the y-intercept of g(x).
Step-by-step explanation:
the highest power of x in f(x) is 3, and it has a positive sign. so, the bigger x the more the x³ part will dominate.
therefore, with x going to positive infinity, also f(x) goes to positive infinity.
the graph of g(x) clearly dives down never to come back up or flatten out. so, with x going to positive infinity, g(x) goes to negative infinity.
the y-intercept of f(x) = f(0).
that means the functional value when x=0.
so, for x=0 this means
y = 2×0³ - 3×0² + 5×0 + 7 = 7
the graph of g(x) shows that g(0) is 8.
so f(0) is less than g(0).
A
Since T is the midpoint of PQ then PT = TQ, hence
6x + 4 = 52 ( subtract 4 from both sides )
6x = 48 ( divide both sides by 6 )
x = 8 → A
Answer:
x = 52.2 degrees
Step-by-step explanation:
In this question, we are to
calculate the value of the angle x
Now the first thing to identify is that what we have is a right angled triangle as one of the angles is 90 degrees
This means that we can employ the use of trigonometric identities to calculate whatever we want to calculate.
The question now is which trigonometric identity is the correct one to use
To answer this, we need to be sure of the sides we were given. Looking at the diagram we have one side facing the 90 degrees angle and the other side facing the angle x itself.
The one facing the angle given is the opposite while the one facing the angle 90 is the hypotenuse
So the trigonometric identity to use is the one that links the opposite and the hypotenuse
This is the sine
mathematically;
sine of an angle = length of opposite/length of hypotenuse
In this case
sine x = 6.4/8.1
sine x = 0.7901
Thus;
x = arc sin 0.7901
x = 52.2 degrees
It would be 19x because it is the same as 19 times x.
Answer:
BD and BE are tangents to the circle
Step-by-step explanation:
Given
See attachment
Required
What does BD and BE represent?
From the attachment, we have:



The measure of angles between a tangent and the radius of a circle is always 
So, because
and AD is a radius, then BD is a tangent to circle.
Similarly,;
and AE being a radius implies that BE is a tangent to circle.