All categories

4 years ago

I need help simplify 1/x^-4 please

Mathematics

1 answer:

Ymorist [56]4 years ago

8 0

X^4 is the simplified answer

You might be interested in

What's 77/88 simplest form

Tems11 [23]

7/8. You can divide each by 11 and get 7/8 which is the simplest form.

3 0

3 years ago

Find the perimeter of a square measuring 5.35 cm on a side.

Alik [6]

Perimeter = 5.35 * 4 = 21.4 cm

5 0

4 years ago

Read 2 more answers

Answer to de best of yo abilities <3

balu736 [363]

So 7.5 is the answer

4 0

3 years ago

How to multiply workout<br><br><br>

DENIUS [597]

Answer:

Steps to multiply using Long Multiplication

Multiplying 2-Digit by 2-Digit Numbers

Let us multiply 47 by 63 using the long multiplication method.

1. Write the two numbers one below the other as per the places of their digits. Write the bigger number on top and a multiplication sign on the left. Draw a line below the numbers.

2. Multiply ones digit of the top number by the ones digit of the bottom number.

Write the product as shown.

3. Multiply the tens digit of the top number by the ones digit of the bottom number.

This is our first partial product which we got on multiplying the top number by the ones digit of the bottom number.

4. Write a 0 below the ones digit as shown. This is because we will now be multiplying the digits of the top number by the tens digit of the bottom number. Hence, we write a 0 in the ones place.

5. Multiply the ones digit of the top number by the tens digit of the bottom number.

6. Multiply the tens digit of the top number by the tens digit of the bottom number.

This is the second partial product obtained on multiplying the top number by the tens digit of the bottom number.

7. Add the two partial products.

In long multiplication method, the number on the top is called the multiplicand. The number by which it is multiplied, that is, the bottom number is called the multiplier.

So, a long division problem will have:

We follow the same method for multiplying numbers greater than 2-Digits.

The figure below shows the long division method to multiply 357 by 23

4 0

3 years ago

How can you prove that csc^2(θ)tan^2(θ)-1=tan^2(θ)

Oxana [17]

Answer:

Make use of the fact that as long as $\sin(\theta) \ne 0$ and $\cos(\theta) \ne 0$ :

$\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ .

$\displaystyle \csc(\theta) = \frac{1}{\sin(\theta)}$ .

$\sin^{2}(\theta) + \cos^{2}(\theta) = 1$ .

Step-by-step explanation:

Assume that $\sin(\theta) \ne 0$ and $\cos(\theta) \ne 0$ .

Make use of the fact that $\tan(\theta) = (\sin(\theta)) / (\cos(\theta))$ and $\csc(\theta) = (1) / (\sin(\theta))$ to rewrite the given expression as a combination of $\sin(\theta)$ and $\cos(\theta)$ .

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \left(\frac{1}{\sin(\theta)}\right)^{2} \, \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} - 1 \\ =\; & \frac{\sin^{2}(\theta)}{\sin^{2}(\theta)\, \cos^{2}(\theta)} - 1\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1\end{aligned}$ .

Since $\cos(\theta) \ne 0$ :

$\displaystyle 1 = \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)}$ .

Substitute this equality into the expression:

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1 \\ =\; & \frac{1}{\cos^{2}(\theta)} - \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}$ .

By the Pythagorean identity, $\sin^{2}(\theta) + \cos^{2}(\theta) = 1$ . Rearrange this identity to obtain:

$\sin^{2}(\theta) = 1 - \cos^{2}(\theta)$ .

Substitute this equality into the expression:

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}$ .

Again, make use of the fact that $\tan(\theta) = (\sin(\theta)) / (\cos(\theta))$ to obtain the desired result:

$\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\\ =\; & \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} \\ =\; & \tan^{2}(\theta)\end{aligned}$ .

5 0

2 years ago