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lilavasa [31]
4 years ago
9

Convert the expression to radical form. 8(1/2)x(1/2)y(3/4) A) 8xy3 B) 2xy 2xy2 C) 2x2y3 D) 2 4x2y3

Mathematics
1 answer:
UNO [17]4 years ago
7 0
Your answer should be B : ) 




I hope this helps!





please mark this as the brainliest 
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1 less than the square of a number.
julsineya [31]

Answer: 1 is not a perfect square. 3 is the only prime number one less than a square.

7 0
3 years ago
Which of the following statements best describes the solution to the problem below?
ZanzabumX [31]

Answer:

C

Step-by-step explanation:

Using rule of negatives and positives:
The rules for multiplication and division are simple: If both numbers are positive, the result is positive. If both numbers are negative, the result is positive. If one number is positive and the other is negative, the result is negative.

This is why C is the correct statement

4 0
2 years ago
Preston drew a blueprint of an additional room he is planning to add to his house. On the blueprint, 1 centimeter is equal to 4
Yuri [45]
It would be 3 cm by 4 cm because you just divide the number of feet by 4.
7 0
3 years ago
The question is on the picture
stich3 [128]

A and C

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6 0
4 years ago
Find the integration of (1-cos2x)/(1+cos2x)
slega [8]

Given:

The expression is:

\dfrac{1-\cos 2x}{1+\cos 2x}

To find:

The integration of the given expression.

Solution:

We need to find the integration of \dfrac{1-\cos 2x}{1+\cos 2x}.

Let us consider,

I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx

I=\int \dfrac{2\sin^2x}{2\cos^2x}dx         [\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]

I=\int \dfrac{\sin^2x}{\cos^2x}dx

I=\int \tan^2xdx                      \left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]

It can be written as:

I=\int (\sec^2x-1)dx             [\because 1+\tan^2 \theta =\sec^2 \theta]

I=\int \sec^2xdx-\int 1dx

I=\tan x-x+C

Therefore, the integration of \dfrac{1-\cos 2x}{1+\cos 2x} is I=\tan x-x+C.

8 0
3 years ago
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