The field hockey team is holding several fundraisers to save up enough money to buy team jackets.

Mathematics

1 answer:

azamat3 years ago

7 0

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Help please I need the name of the polygon and the perimeter and area of the shape of numbers 42-47

sergij07 [2.7K]

42 - 47

Polygons are

triangle, trapezoid, triangle, parallelogram, rectangle and parallelogram

The area of triangles = 1/2 * base * haight.

Area of the parallelograms = height * base.

Area of rectangle= width * length

Area of trapezoid = 1/2 * (a + b) * height where a and b are the opposite parallel sides

The perimeters = sum of the sides of the polygons.

8 0

2 years ago

Let a and b be roots of x² - 4x + 2 = 0. find the value of a/b² +b/a²

erastovalidia [21]

Answer:

$\dfrac{a}{b^2}+\dfrac{b}{a^2}=10$

Step-by-step explanation:

Given equation: $x^2-4x+2=0$

The roots of the given quadratic equation are the values of x when $y=0$ .

To find the roots, use the quadratic formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=1, \quad b=-4, \quad c=2$

$\begin{aligned}\implies x & =\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}\\& =\dfrac{4 \pm \sqrt{8}}{2}\\& =\dfrac{4 \pm 2\sqrt{2}}{2}\\& =2 \pm \sqrt{2}\end{aligned}$

$\textsf{Let }a=2+\sqrt{2}$

$\textsf{Let }b=2-\sqrt{2}$

Therefore:

$\begin{aligned}\implies \dfrac{a}{b^2}+\dfrac{b}{a^2} & = \dfrac{2+\sqrt{2}}{(2-\sqrt{2})^2}+\dfrac{2-\sqrt{2}}{(2+\sqrt{2})^2}\\\\& = \dfrac{2+\sqrt{2}}{6-4\sqrt{2}}+\dfrac{2-\sqrt{2}}{6+4\sqrt{2}}\\\\& = \dfrac{(2+\sqrt{2})(6+4\sqrt{2})+(2-\sqrt{2})(6-4\sqrt{2})}{(6-4\sqrt{2})(6+4\sqrt{2})}\\\\& = \dfrac{12+8\sqrt{2}+6\sqrt{2}+8+12-8\sqrt{2}-6\sqrt{2}+8}{36+24\sqrt{2}-24\sqrt{2}-32}\\\\& = \dfrac{40}{4}\\\\& = 10\end{aligned}$