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Vesna [10]
4 years ago
11

Here are the first five terms of a sequence.

Mathematics
1 answer:
Levart [38]4 years ago
8 0

Answer:

So the formula for the sequence is

Second term =first term + (a+nd)

Third term = second term + (a+nd)

And so on

Where n start with zero

Step-by-step explanation:

3 8 17 30 47

First term = 3

Second term = 8

Third term = 17

Fourth term = 30

Fifth term = 47

Second term - first term = 8-3=5

Third term - second term = 17-8= 9

Fourth term - third term = 30-17=13

Fifth term - fourth term = 47-30=17

So the AP is 5 9 13 17

We find out that the common difference is 9-5 = 4

The formula for nth term of the AP is

= a + nd

So the formula fir the sequence is

Second term =first term + (a+nd)

Third term = second term + (a+nd)

And so on

Where n start with zero

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Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as:  A1=\frac{(base)(height)}{2}

Area of a rectangle is calculated as: A2=(side)(side)

Area of a right trapezoid is: A3=\frac{(a+b)h}{2}, where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

S1=\frac{(2x-3)(4x-6)}{2}

S1=4x^{2}-12x+9

Area of rectangle S2:

S2 = (4x-6)(3x-2)

S2=12x^{2}-26x+12

Area of trapezoid S3:

S3=\frac{(2x+3+4x+1)(2x-3)}{2}

S3=\frac{(6x+4)(2x-3)}{2}

S3=6x^{2}-5x-6

2) a) S=4x^{2}-12x+9+12x^{2}-26x+12-(6x^{2}-5x-6)

S=4x^{2}-12x+9+12x^{2}-26x+12-6x^{2}+5x+6

S=10x^{2}-33x+37

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

\frac{(3x-2+3x-2+2x-3)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}

\frac{(8x-7)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}

32x^{2}-48x-28x+42=12x^{2}+8x-18x-12

20x^{2}-66x+54=0

Using Bhaskara to solve the second degree equation:

\frac{66+\sqrt{(-66)^{2}-(4.20.54)} }{2(20)}

x_{1}=\frac{66+6}{40} = 1.8

x_{2}=\frac{66-6}{40} = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) <u>Expand</u> <u>a</u> <u>polynomial</u> (or equation) is to multiply all the terms, remiving the parenthesis. <u>Reduce</u> <u>a</u> <u>polynomial</u> (or equation) is to combine terms alike,e.g.:

S=(2x-3)(5x-9)

S=10x^{2}-18x-15x+27 (expand)

S=10x^{2}-33x+27 (reduce)

d) For area of AFCG to be bigger than area of ADEB by 27:

32x^{2}-48x-28x+42=12x^{2}+8x-18x-12+27

32x^{2}-48x-28x+42=12x^{2}+8x-18x+15

20x^{2}-66x+27=0

Solving:

\frac{66+\sqrt{(-66)^{2}-(4.20.27)} }{2(20)}

\frac{66+46.86}{40}

x_{1}=\frac{66+46.86}{40}= 2.82

x_{2}=\frac{66-46.86}{40} = 0.48

According to the enunciation, x cannot be less than 1.5, then, the value of x so that area AFGC exceeds the area ADEB by 27 is 2.82

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3 years ago
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