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4 years ago

Find the explicit formula for the arithmetic sequence -31,-27,-23,-19 b(n)= ?

Mathematics

2 answers:

BigorU [14]4 years ago

4 0

b(n)=-31+4(n-1) is the explicit formula

PolarNik [594]4 years ago

4 0

$a_n=a_1+(n-1)d\\ a_1=-31\\ d=a_n-a_{n-1}=-27-(-31)=-27+31=4\\ a_n=-31+(n-1)\cdot 4\\a_n=-31+4n-4\\ a_n=4n-35$

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4 years ago

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Use the given figure to find the length of FG

ra1l [238]

ANSWER

$|FG| = 19 \: units$

EXPLANATION

Since the altitude, |GJ| bisects the base of the triangle, it follows that, the triangle is an isosceles triangle.

This implies that,.

$|FG|=|GH|$

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$5x - 3x = 4 + 6$

This implies that,

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3 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

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csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$