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Ede4ka [16]
3 years ago
8

2(x-3)^2+10=82 solve by isolating x

Mathematics
1 answer:
Leviafan [203]3 years ago
7 0
We have the following equation:
 2 (x-3) ^ 2 + 10 = 82
 We must clear x.
 We pass the constant terms to one side of the equation:
 2 (x-3) ^ 2 = 82 - 10
 2 (x-3) ^ 2 = 72
 (x-3) ^ 2 = 72/2
 (x-3) ^ 2 = 36
 Square root to both members:
 x-3 = +/- root (36)
 x-3 = +/- 6
 The solutions are:
 x1 = 6 + 3 = 9
 x2 = -6 + 3 = -3
 Answer:
 
x1 = 9
 
x2 = -3
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Answer:

The area of rhombus PQRS is 120 m.

Step-by-step explanation:

Consider the rhombus PQRS.

All the sides of a rhombus are equal.

Hence, PQ = QR = RS = SP = 13 m

The diagonals PR and QS bisect each other.

Let the point at of intersection of the two diagonals be denoted by <em>X</em>.

Consider the triangle QXR.

QR = 13 m

XR = 12 m

The triangle QXR is a right angled triangle.

Using the Pythagorean theorem compute the length of QX as follows:

QR² = XR² + QX²

QX² = QR² - XR²

       = 13² - 12²

       = 25

 QX = √25

       = 5 m

The measure of the two diagonals are:

PR = 2 × XR = 2 × 12 = 24 m

QS = 2 × QX = 2 × 5 = 10 m

The area of a rhombus is:

\text{Area}=\frac{1}{2}\times d_{1}\times d_{2}

Compute the area of rhombus PQRS as follows:

\text{Area}=\frac{1}{2}\times PR\times QS

        =\frac{1}{2}\times 24\times 10\\\\=120

Thus, the area of rhombus PQRS is 120 m.

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