Let abc be an isosceles triangle with ab = ac and ∠bac=30°. side ab and the median ad contain points q and p respectively (points p and q are not at point a) in such a way that pc=pq. find m∠pqc.

  15°

Step-by-step explanation:

Since P is on the median of ΔABC, it is equidistant from points B and C as well as from C and Q. Thus, points B, C, and Q all lie on a circle centered at P. (See the attached diagram.)

The base angles (B and C) of triangle ABC are (180° -30°)/2 = 75°. This means arc QC of the circle centered at P has measure 150°. The diameter of circle P that includes point Q is defined to intersect circle P at R.

Central angle RPC is the difference between arcs QR and QC, so is 180° -150° = 30°. Inscribed angle RQC has half that measure, so is 15°.  Angle PQC has the same measure as angle RQC, so is 15°.

Angle PQC is 15°.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

what graph

step-by-step explanation: