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2 years ago

Simplify, Write your answer in exponential notation. a.) 2³ x 2-5 b.) (602)3

Mathematics

1 answer:

mariarad [96]2 years ago

8 0

The simplified expressions are 2^-2 and 60^6

<h3>How to simplify the expressions?</h3>

Expression (a)

We have:

2^3 * 2^-5

Apply the product law of indices

2^3 * 2^-5 = 2^(3 - 5)

Evaluate the difference

2^3 * 2^-5 = 2^-2

Expression (b)

We have:

(60^2)^3

Apply the power law of indices

(60^2)^3 = 60^(2 * 3)

Evaluate the product

(60^2)^3 = 60^6

Hence, the simplified expressions are 2^-2 and 60^6

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Step-by-step explanation:

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2 years ago

What is the simplified form of each expression? 7x^–8 × 6x^3

klio [65]

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3 years ago

Find sin X, sin Z, cos X, and cos Z. Write each answer as a simplified fraction.

just olya [345]

Answer:

Find the answers below

Step-by-step explanation:

Using m<X as the reference angle

Opposite YZ = 7

Adjacent XY = 10

Hypotenuse XZ = √149

Using the SOH CAH TOA identity

sinX = opp/hyp

sinX =YZ/XZ

sinX = 7/√149

For cos X

cos X = adj/hyp

cos X =10/√149

Using m<Z as reference angle;

Opposite XY = 10

Adjacent YZ = 7

Hypotenuse XZ = √149

Using the SOH CAH TOA identity

sinZ = opp/hyp

sinZ =10/√149

sinZ = 7/√149

For cos Z

cosZ = 7/√149

7 0

2 years ago

find the Taylor Series for tan−1(x) based at 0. Give your answer using summation notation and give the largest open interval on

enyata [817]

Let $f(x)=\tan^{-1}x$ . Then $f'(x)=\frac1{1+x^2}$ . Note that $f(0)=0$ .

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n\ge0}x^n$

This means that for $|-x^2|=|x|^2$ , or $-1$ , we have

$\displaystyle f'(x)=\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n\ge0}(-x^2)^n=\sum_{n\ge0}(-1)^nx^{2n}$

Integrate the series to get

$f(x)=f(0)+\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

$\implies\tan^{-1}x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

6 0

3 years ago