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Cerrena [4.2K]
1 year ago
12

What are the solutions of x² + 6x - 16 = 0?

Mathematics
1 answer:
AveGali [126]1 year ago
6 0

Step-by-step explanation:

x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \left(-16\right)}}{2\cdot \:1}\\\sqrt{6^2-4\cdot \:1\cdot \left(-16\right)}\\\sqrt{6^2+64}\\\sqrt{36+64}\\\sqrt{100}\\\sqrt{10^2}\\=10\\x_1=\frac{-6+10}{2\cdot \:1},\:x_2=\frac{-6-10}{2\cdot \:1}\\\\\bold{x_1:2}\\\frac{-6+10}{2\cdot \:1}\\\frac{4}{2\cdot \:1}\\\frac{4}{2}\\=2\\\\\bold{x_2:-8}\\\frac{-6-10}{2\cdot \:1}\\\frac{-16}{2\cdot \:1}\\\frac{-16}{2}\\-\frac{16}{2}\\=-8

Answer:

x₁ = 2

x₂ = -8

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In an isosceles trapezoid the length of a diagonal is 25 cm and the length of an altitude is 15 cm. Find the area of the trapezo
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Answer:

The area of the trapezoid is 525\ cm^{2}

Step-by-step explanation:

we know that

The area of a isosceles trapezoid is equal to the area of two isosceles right triangles plus the area of a rectangle

step 1

<em>Find the area of the  isosceles right triangle</em>

Remember that

In a isosceles right triangle the height is equal to the base of the triangle

we have

h=15\ cm

so

b=15\ cm

The area is equal to

A=\frac{1}{2}(b)(h)

substitute the values

A=\frac{1}{2}(15)(15)=112.5\ cm^{2}

step 2

Find the area of the rectangle

The area of the rectangle is equal to

A=LW

we have

W=15\ cm -----> is the height of the trapezoid

d=25\ cm  -----> the diagonal of the rectangle

Applying the Pythagoras Theorem

25^{2}=L^{2}+15^{2}\\L^{2}=25^{2}-15^{2} \\ L^{2} =400\\L=20\ cm

The area of the rectangle is

A=(20)(15)=300\ cm^{2}

step 3

Find the area of the trapezoid

A=2(112.5\ cm^{2})+300\ cm^{2}=525\ cm^{2}

6 0
4 years ago
Read 2 more answers
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