A balalaika is a Russian stringed instrument. Show that the triangular parts of the two balalaikas are congruent for x = 6.

*the diagram of the Russian stringed instrument is attached below.

Answer/Step-by-step explanation:

To show that the traingular parts of the two balalaikas instruments are congruent, substitute x = 6, to find the missing measurements that is given in both ∆s.

Parts of the first ∆:

WY = (2x - 2) in = 2(6) - 2 = 12 - 2 = 10 in

m<Y = 9x = 9(6) = 54°.

XY = 12 in

Parts of the second ∆:

m<F = 72°

HG = (x + 6) in = 6 + 6 = 12 in

HF = 10 in

m<G = 54°

m<H = 180 - (72° + 54°)

m<H = 180 - 126

m<H = 54°

From the information we have, let's match the parts that are congruent to each other in both ∆s:

WY ≅ FH (both are 10 in)

XY ≅ GH (both are 12 in)

<Y ≅ <G (both are 54°)

Thus, since two sides (WY and XY) and an included angle (<Y) of ∆WXY is congruent to two corresponding sides (FH and GH) and an included angle (<G) in ∆FGH, therefore, ∆WXY ≅ ∆FGH by the Side-Angle-Side (SAS) Congruence Theorem.

This is enough proof to show that the triangular parts of the two balalaikas are congruent for x = 6.

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.

addition

step-by-step explanation: