CAN SOMEONE HELP ME WITH THIS PLEASE ??
1 answer:
Remark
The very first thing you should do is graph the function, which I have done. It is at the bottom of this answer.
Step One
Where is the hole and what kind is it? This graph has a hole at x = 2. At that point the function is 0/0. A hole is defined as the point created by a zero in one of the factors that is the same in the numerator and denominator. Both = 0.
Step 2
Find the vertical asymptote.
That occurs when denominator alone has a factor that is = 0. In this case when x = 3 the whole equation blows up (or down) and you have a vertical asymptote. The graph is discontinuous at this point.
Step three
Find the horizontal Asymptote.
Horizontal assymptotes occur when (in this case) the power of the factors in the denominator exceeds the power of the factors in the numerator. In this case that is true and the horizontal assymptote occurs around the line y = 0.
The very first thing you should do is graph the function, which I have done. It is at the bottom of this answer.
Step One
Where is the hole and what kind is it? This graph has a hole at x = 2. At that point the function is 0/0. A hole is defined as the point created by a zero in one of the factors that is the same in the numerator and denominator. Both = 0.
Step 2
Find the vertical asymptote.
That occurs when denominator alone has a factor that is = 0. In this case when x = 3 the whole equation blows up (or down) and you have a vertical asymptote. The graph is discontinuous at this point.
Step three
Find the horizontal Asymptote.
Horizontal assymptotes occur when (in this case) the power of the factors in the denominator exceeds the power of the factors in the numerator. In this case that is true and the horizontal assymptote occurs around the line y = 0.
You might be interested in
Answer:
Option B is correct.
Step-by-step explanation:
We have given a triangle ABC and EDC please look at the figure
We can see that AE and BD are transversal therefore, ∠EAB=∠AED being alternate interior angles
And ∠ACB=∠DCE are vertically opposite angles hence, equal
So, by AA similarity postulate the above to triangles are similar
ΔABC ΔEDC
Therefore, Option B is correct that is Triangle ABC is similar to triangle EDC , because m∠3 = m∠4 and m∠1 = m∠5
NOTE: m∠3 = m∠4 corresponds to m∠ACB=m∠DCE
And m∠1 = m∠5 corresponds to m∠EAB=m∠AED
Answer:
AC = { 4, 5, 6, 7 }
Step-by-step explanation:
If you see, the diagonal AC forms two triangles, Δ ABC, and Δ ADC. In Δ ABC, AC = 3 units and BC = 6 units, while AC is yet to be known. Respectively in Δ ADC, AD = 4 units and CD = 4 units, while AC is again yet to be known.
In both triangles the triangle inequality can help find the possible value( s ) of AD, as this inequality only restricts some of the possible values with which AC can take. At the same time AC is shared among the two triangles, so if we can apply the Triangle Inequality to both of these triangles, the value of AC can be " further restricted. "
And there we have two inequalities, 3 < AC < 9, and 0 < AC < 8. Combining both inequalities the only possible integer values for AC would be 4, 5, 6, and 7.