+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+9g(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 ∠AHC and ∠AOC is 240 by the given info. Notice that m∠AOC=m∠AHC because they both intercept the same arc. Therefore, we can solve for the angle of the circumcenter.
| m∠AOC+m∠AHC=240 | Since we already established that m∠AOC=m∠AHC, we can solve for the the central angle of the circle. |
| m∠AOC+m∠AOC=240 | |
| 2m∠AOC=240 | |
| m∠AOC=120∘ | There is no need to solve for the angle measure of the orthocenter; it is simply a waster of time. |
Notice that ∠ABC is an inscribed angle, which means that its measure is equal to half the measure of the central angle.
| 12m∠AOC=m∠ABC | Use substitution here since we already know the measure of one of the angles. |
| 12∗120=m∠ABC | |
| m∠ABC=60∘ |
+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)
| △ABC is an isosceles right trianglem∠A=90∘O is the circumcenter | Given information |
| m∠A+m∠B+m∠C=180∘ | Triangle sum theorem (the sum of the interior angles of a triangle equals 180 degrees) |
| 90∘+m∠B+m∠C=180∘ | Substitution property of equality |
| m∠B+m∠C=90∘ | Subtraction property of equality |
| m∠B=m∠C | Isosceles Triangle Theorem (The angles opposite congruent sides in an isosceles triangle are congruent) |
| m∠B+m∠B=90∘ | Substitution property of equality |
| 2m∠B=90∘ | Simplify |
| m∠B=45∘ | Division property of equality |
| ¯AO≅¯BO | Property of Circumcenter (The circumcenter is equidistant from the vertices of the triangle) |
| m∠A=m∠B | |
Well, this is taking some time...
Anyway, △AOB is another isosceles triangle. m∠OAB=m∠B=45∘, so the remaining angle, ∠BOA=90∘
.
+118704 Hello DH,
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