Log²(x + 5) + log²(x - 5) = log²11

Complete the missing parts of the paragraph proof. Draw a perpendicular from P to AB. Label the intersection C. We are given tha

zvonat [6]

<span>1) We are given that PA = PB, so PA ≅ PB by the definition of the radius.

</span>When you draw a perpendicular to a segment AB, you take the compass, point it at A and draw an arc of size AB, then you do the same pointing the compass on B. Point P will be one of the intersections of those two arcs. Therefore PA and PB correspond to the radii of the arcs, which were taken both equal to AB, therefore they are congruent.

2) We know that angles PCA and PCB are right angles by the definition of perpendicular.

Perpendicularity is the relation between two lines that meet at a right angle. Since we know that PC is perpendicular to AB by construction, ∠PCA and ∠PCB are right angles.

3) PC ≅ PC by the reflexive property congruence.

The reflexive property congruence states that any shape is congruent to itself.

4) So, triangle ACP is congruent to triangle BCP by HL, and AC ≅ BC by CPCTC (corresponding parts of congruent triangles are congruent).

CPCTC states that if two triangles are congruent, then all of the corresponding sides and angles are congruent. Since ΔACP ≡ ΔBCP, then the corresponding sides AC and BC are congruent.

5) Since PC is perpendicular to and bisects AB, P is on the perpendicular bisector of AB by the definition of the perpendicular bisector.

<span>The perpendicular bisector of a segment is a line that cuts the segment into two equal parts (bisector) and that forms with the segment a right angle (perpendicular). Any point on the perpendicular bisector has the same distance from the segment's extremities. PC has exactly the characteristics of a perpendicular bisector of AB. </span>

6 0

3 years ago

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Can I get help on this please? I think the answer might be (A) but i'm not sure.

Alex Ar [27]

Yeah it would be A. because x approaching infinity is to the right of the graph, and since there is a horizontal asymptote at -3 then its correct

4 0

3 years ago

The area of a rectangular patio is 31 7/8 square meters. The width of the patio is 4 1/4 meters. What is the length? ( in simple

Allushta [10]

Answer:

it is good and people need to do math problems

Step-by-step explanation:

8 0

2 years ago

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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$