In δpqr shown below, segment qs is an altitude: triangle pqr with segment sq drawn from vertex q and intersecting side rp. which of the following is a justification used while proving the similarity of triangles δpsq and δqsr? a. transitive property of equality b. addition property of equality c. definition of an altitude d. definition of supplementary angles

c. Definition of altitude.

Step-by-step explanation:

We are given that segment QS is an altitude in ΔPQR and we are asked to find a justification used while proving the similarity of triangles ΔPSQ and ΔQSR.

Since we know that altitude meets opposite side at right angles. When QS will intersect line PR we will get two right triangles QSR and QSP right angled at S.

ΔPQR is similar to ΔPSQ as they both share angle P and right triangle. So their third angle should also be similar.

ΔPQR is similar to ΔQSR as they both share angle R and both have a right triangle at Q and S respectively. So they will have their third angle equal.

ΔPQR is similar to triangles ΔQSR and ΔPSQ. Therefore, ΔQSR is similar to ΔPSQ.

Therefore, by definition of altitude triangles ΔPSQ and ΔQSR are similar as ΔPSQ and ΔQSR are created from ΔPQR by altitude QS.

a function could be identified in 3 ways either on a chart,graph, or ordered pairs. first a function is that each point has its own. the rule is that there can be many x's to one y but there can not be 2 x's at different y's. so on a graph when you put a straight line through the y's there should be one dot only to be a function.

step-by-step explanation:

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 =

x = 3, y = 216 =

x = 4, y = 1296 =

x = 5, y = 7776 =

so, the geometric series obtained is .

thus, the common ratio is given by = = 6.

hence, the common ratio is 6.