3 years ago

Find a formula for the nth term in this arithmetic sequence. a1=0, a2=0.5, a3=1, a4=1.5

Mathematics

1 answer:

MrRissso [65]3 years ago

6 0

Answer:

nth term is;

0.5n-0.5

Step-by-step explanation:

As we can see, the first term is 0

The common difference is 0.5-0= 1-0.5 = 1.5-1 = 0.5

Formula for nth term of an arithmetic sequence is;

a + (n-1)d

So we have

0 + (n-1) 0.5

= 0.5n-0.5

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3 0

3 years ago

Read 2 more answers

Which expression is equivalent to square root 55x^7y6/11x^11y^8

Novay_Z [31]

Answer:

$\large\boxed{\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\dfrac{\sqrt5}{x^2y}}$

Step-by-step explanation:

$\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{55}{11}\cdot\dfrac{x^7}{x^{11}}\cdot\dfrac{y^6}{y^8}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=\sqrt{5x^{7-11}y^{6-8}}=\sqrt{5x^{-4}y^{-2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt5\cdot\sqrt{x^{-4}}\cdot\sqrt{y^{-2}}=\sqrt5\cdot\sqrt{x^{(-2)(2)}}\cdot\sqrt{y^{(-1)(2)}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt5\cdot\sqrt{(x^{-2})^2}\cdot\sqrt{(y^{-1})^2}\qquad\text{use}\ \sqrt{a^2}=a$

$=\sqrt5\cdot x^{-2}\cdot y^{-1}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\to a^{-1}=\dfrac{1}{a}\\\\=\sqrt5\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}=\dfrac{\sqrt5}{x^2y}$