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1 answer:

7 0

The original square has the area (8 in)^2 = 64 in^2.

If we mult. this area by 36, we get the area of a larger square 2304 in^2.

The new side length is sqrt(2304 in^2), or 48. In other words, the original square has side length 8 in, but the 'new' square has side length 48 in.

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Consider the quadratic function.

Genrish500 [490]

Mark Brainliest please

Answer :

a = 1, b = -8, c = -5

Explanation

A standard form for the quadratic function is f(x) = ax2 + bx + c where the coefficients are a, b, and c.
The given quadratic function is f(p) = P2 - 8P - 5 When this function is compared to the standard form, with p replacing x, we obtain a = 1 because the leading term is 1*p2, b = -8 because the linear term is -8*p, c = -5 because the constant term is -5.
Answer is : a = 1, b = -8, c = -5.

6 0

3 years ago

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

Ann [662]

*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.