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Mazyrski [523]
3 years ago
15

ASAP plz answers just

Mathematics
1 answer:
faust18 [17]3 years ago
8 0

Answer:

<em>(a) x(x+2)  (b) 2x(x-3)  (c) 5x(3-2x^2)  (d) 3x^2(3+x)</em>

Step-by-step explanation:

For each expression, <u>we need to factor out the GCF(Greatest common factor)</u>

a)   x^2+2x

Here, the GCF is  x. So, we will get.....

x^2+2x\\ \\ =x(\frac{x^2}{x}+ \frac{2x}{x}) \\ \\ =x(x+2)

b)   2x^2-6x

Here, the GCF is  2x. So, we will get......

2x^2-6x\\ \\ =2x(\frac{2x^2}{2x}- \frac{6x}{2x}) \\ \\ =2x(x-3)

c)   15x-10x^3

Here, the GCF is  5x. So, we will get.......

15x-10x^3\\ \\ =5x(\frac{15x}{5x}- \frac{10x^3}{5x}) \\ \\ =5x(3-2x^2)

d)   9x^2+3x^3

Here, the GCF is  3x^2. So, we will get.......

9x^2+3x^3\\ \\ =3x^2(\frac{9x^2}{3x^2}+ \frac{3x^3}{3x^2}) \\ \\ =3x^2(3+x)

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I need help with this problem. Given: BC = 10 inches AC = √50 inches m∠CBD = 60° m∠CAD = 90° Calculate the exact area of the sha
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Answer:

Area of the shaded region = 23.33 in²

Step-by-step explanation:

Area of a sector = \frac{\theta}{360}(\pi r^{2})

Where θ = Central angle subtended by an arc

r = radius of the circle

Area of the sector BCD = \frac{60}{360}(\pi) (10^{2})

                                       = 52.36 in²

Area of equilateral triangle BCD = \frac{\sqrt{3} }{4}(\text{Side})^2

                                                      = \frac{\sqrt{3} }{4}(10)^2

                                                      = 25\sqrt{3} in²

                                                      = 43.30 in²

Area of the shaded portion in ΔBCD = 52.36 - 43.3

                                                             = 9.06 in²

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                                 = 39.27 in²

Area of right triangle CAD = \frac{1}{2}(\text{Base})(\text{Height})

                                            = \frac{1}{2}(\text{AC})(\text{AD})

                                            = \frac{1}{2}(\sqrt{50})(\sqrt{50})

                                            = 25 in²

Area of the shaded part in the ΔACD = 39.27 - 25

                                                                         = 14.27 in²

Area of the shaded part of the figure = 9.06 + 14.27

                                                                = 23.33 in²

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