(x^2+3) - [3x+(8-x^2) ]

Mathematics

1 answer:

Maksim231197 [3]3 years ago

7 0

$2 {x}^{2} - 3x - 5$ ✅

Step-by-step explanation:

$( {x}^{2} + 3) - [3x + (8 - {x}^{2} )] \\ = {x}^{2} +3 - [3x + 8 - {x}^{2} ] \\ = {x}^{2} + 3 - 3x - 8 + {x}^{2} \\ = 2 {x}^{2} - 3x - 5$

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Answer:

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kykrilka [37]

Hey!
So you have a net of the figure. It's broken into several 2D shapes. What you an do it find the area of each shape and add it together.
Triangles:
A = (b*h)/2
A= (5 * 3) / 2
A = 15/2
same thing for the second one
15/2 * 2 = 15
Don't forget, there are 2 triangles so it's really 15 for both

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A=b*h
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A=7*5
A=35

A=3*7
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3 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

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The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by: