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3 years ago

How do you simplify the exponent, p* p²

2 answers:

8 0

Firstly, I would recommend that you learn all the exponent laws :)
For this question, we will actually be using one of these laws!

This law can be only used when two or more numbers/ variables with exponents are multiplied together.

Simply add the exponents together to simplify.
In this case:

p x p²= p³

(the exponents 1 and 2 add to 3)

Hope this helps!
Feel free to message me if you have any questions :)

kogti [31]3 years ago

4 0

P*p^2 :Exponent:

= p^3 :Solution:

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3 years ago

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3 years ago

Find the value of the variable and yz if y is between x and z Xy=6b yz=8b Xz=175

Anika [276]

The value of b is 12.5 and the value of YZ is 100

<h3>How to determine the variable and yz?</h3>

The given parameters are:

XY = 6b

YZ = 8b

XZ = 175

This means that

XZ = XY + YZ

So, we have:

6b + 8b = 175

Evaluate the sum

14b = 175

Divide by 14

b = 12.5

Substitute b = 12.5 in YZ = 8b

YZ = 8 * 12.5

Evaluate

YZ = 100

Hence, the value of b is 12.5 and the value of YZ is 100

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brainly.com/question/24571594

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2 years ago

If k stands for an integer, then is it possible for k2 + k to stand for an odd integer? Be prepared to justify your answer.

BartSMP [9]

Answer:

<em>k^2 + k never stands for an odd integer</em>

Step-by-step explanation:

Let us consider either case with which k stands for an odd or even integer;

Case 1: k is an odd integer

For integer a, k = 2a + 1

So, k + 1 = 2a + 2 = 2( a + 1 ) = 2b for integer b

k^2 + k = k ( k + 1 ) = k ( 2b ) = 2kb = 2c for integer c,

<em>Therefore, if k is an odd integer, then k^2 + k is an even integer ;</em>

Case 2: k is an even integer

For an integer a, k = 2a

So, k + 1 = 2a + 1

k^2 + k = k( k+1 ) = 2a( 2a + 1 ) , multiple of 2

<em>Therefore, if k is an even integer, then k^2 + k is an even integer;</em>

<em>This would make k^2 + k never stand for an odd integer</em>

5 0

3 years ago