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2 years ago

the shapes below are drawn on a centimeter grid, show that the triangle and the square have the same area.

Mathematics

1 answer:

guajiro [1.7K]2 years ago

3 0

Answer:

Area_square = 4 cm2

Area_triangle = 4 cm2

Step-by-step explanation:

The first thing to do is calculate the area of both figures.

The area of the square is given by:

Area_square = side * side

The side of the square is 2 cm, so:

Area_square = 2 * 2 = 4 cm2

The area of the triangle is given by:

Area_triangle = base * height / 2

The base is 2 cm, and the height is 4 cm, so we have:

Area_triangle = 2 * 4 / 2 = 4 cm2

So we have that Area_square = Area_triangle = 4 cm2

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Step-by-step explanation:

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2 years ago

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2 years ago

Find a quadratic polynomial with integer coefficients which has x = 2/9, ± SQRT23/9 as its real zeros.

melomori [17]

Answer:

The quadratic polynomial with integer coefficients is $y = 81\cdot x^{2}-36\cdot x -19$ .

Step-by-step explanation:

Statement is incorrectly written. Correct form is described below:

<em>Find a quadratic polynomial with integer coefficients which has the following real zeros: </em> $x = \frac{2}{9}\pm \frac{\sqrt{23}}{9}$ <em>. </em>

Let be $r_{1} = \frac{2}{9}+\frac{\sqrt{23}}{9}$ and $r_{2} = \frac{2}{9}-\frac{\sqrt{23}}{9}$ roots of the quadratic function. By Algebra we know that:

$y = (x-r_{1})\cdot (x-r_{2}) = x^{2}-(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2}$ (1)

Then, the quadratic polynomial is:

$y = x^{2}-\frac{4}{9}\cdot x -\frac{19}{81}$

$y = 81\cdot x^{2}-36\cdot x -19$

The quadratic polynomial with integer coefficients is $y = 81\cdot x^{2}-36\cdot x -19$ .

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2 years ago

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Step-by-step explanation:

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