+2446 The steps are not too long-winded luckily. Unfortunately for Cphill, the 3 that precedes the argument is missing from Cphill's answer.
1) As Cphill mentioned, if \(f(x)\) is the original function, then \(f(x-6)\) translates the function 6 units rightward. The instructions specificially request for the answer with an equation of \(g(x)\) in terms of \(f(x)\), \(g(x)=f(x-6)\)
2) In the previous answer, we determined that \(g(x)=f(x-6)\), so we can use this information to solve for the current problem:
| \(g(x)=f(x-6)\\ f(x)=3(x-5)^2+9\) | \(f(x)=3(x-5)^2+9\) |
| \(f(x-6)=3[(x-6)-5]^2+9\) | |
| \(f(x-6)=3(x-11)^2+9\) | |
| \(g(x)=f(x-6)=3(x-11)^2+9\\ g(x)=3(x-11)^2+9\) |
3) If \(g(x)\) is translated 7 units downward to obtain \(j(x)\), then \(g(x)-7=j(x)\). Of course, the question asks for the equation to be in terms of f(x):
| \(j(x)=g(x)-7\) | Now, substitute g(x)=f(x-6). |
| \(j(x)=f(x-6)-7\) | |
4) Find j(x) in terms of x:
| \(j(x)=g(x)-7\) | We have determined previously that \(g(x)=3(x-11)^2+9\), so substitute that in. |
| \(j(x)=[3(x-11)^2+9]-7\) | Now, simplify. There is not much to do. |
| \(j(x)=3(x-11)^2+2\) | |
+2446 The second one was definitely much harder for me to come up with! Here is a picture that may be useful to reference as I solve:
You will probably notice that I added a circle that inscribes the triangle. At its bost basic level, the circumcenter is a point that is equidistant from all the vertices of the triangle. The circumcenter is also the center of a triangle's circumcircle, which I have illustrated with the diagram above. Of course, we know that the sum of \(\angle AHC\) and \(\angle AOC\) is 240 by the given info. Notice that \(m\angle AOC=m\angle AHC\) because they both intercept the same arc. Therefore, we can solve for the angle of the circumcenter.
| \(m\angle AOC+m\angle AHC=240\) | Since we already established that \(m\angle AOC=m\angle AHC\), we can solve for the the central angle of the circle. |
| \(m\angle AOC+m\angle AOC=240\) | |
| \(2m\angle AOC=240\) | |
| \(m\angle AOC=120^\circ\) | There is no need to solve for the angle measure of the orthocenter; it is simply a waster of time. |
Notice that \(\angle ABC\) is an inscribed angle, which means that its measure is equal to half the measure of the central angle.
| \(\frac{1}{2}m\angle AOC=m\angle ABC\) | Use substitution here since we already know the measure of one of the angles. |
| \(\frac{1}{2}*120=m\angle ABC\) | |
| \(m\angle ABC=60^\circ\) |
+2446 I have been thinking about these two problems for some time, and I think I have finally cracked them! I'll now impart my knowledge to you!
1)
| \(\triangle ABC\text{ is an isosceles right triangle}\\ m\angle A=90^{\circ}\\ O\text{ is the circumcenter}\) | Given information |
| \(m\angle A+m\angle B+m\angle C=180^{\circ}\) | Triangle sum theorem (the sum of the interior angles of a triangle equals 180 degrees) |
| \(90^{\circ}+m\angle B+m\angle C=180^{\circ}\) | Substitution property of equality |
| \(m\angle B+m\angle C=90^\circ\) | Subtraction property of equality |
| \(m\angle B=m\angle C\) | Isosceles Triangle Theorem (The angles opposite congruent sides in an isosceles triangle are congruent) |
| \(m\angle B+m\angle B=90^\circ\) | Substitution property of equality |
| \(2m\angle B=90^\circ\) | Simplify |
| \(m\angle B=45^\circ\) | Division property of equality |
| \(\overline{AO}\cong\overline{BO}\) | Property of Circumcenter (The circumcenter is equidistant from the vertices of the triangle) |
| \(m\angle A=m\angle B\) | |
Well, this is taking some time...
Anyway, \(\triangle AOB\) is another isosceles triangle. \(m\angle OAB=m\angle B=45^\circ\), so the remaining angle, \(\angle BOA=90^\circ\)
.
+118690 Hello DH,
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