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3 years ago

How do i use product rule and simplfy while keeping positive exponents

1 answer:

4 0

Product rule is
$( x^{m} )( x^{n} )= x^{m+n}$
so if ou have
$( x^{2} )( x^{3} )= x^{2+3}=x^{5}$

not sure what yo mean by positive exponents, just add them

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Need help, please<br> Find EF

KonstantinChe [14]

Answer:

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Step-by-step explanation:

8 0

2 years ago

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castortr0y [4]

Answer:

1/2 x -5/2 = -5/4

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4 0

3 years ago

How to reflect a point on a unit circle across axis?

dlinn [17]

Start at the given point. Draw a segment from that point through the center of the circle, and extend the segment until it intersects the circle. The new point of intersection of the segment and the circle is the image of the original point.

7 0

3 years ago

It’s in the picture if u can help I will be thankful

Delicious77 [7]

Answer:

The length of LM,hypotenuse is 55.866ft

Step-by-step explanation:

Given that LN=39ft and MN=40ft so you can solve it using Pythogoras' Theorem:

side²+side² = hypo²

LN² + MN² = LM²

hypo² = 39²+40²

= 3121

LM = √3121

= 55.866ft (3d.p)

6 0

3 years ago

A and b are positive integers and a-b=2. Evaluate the following:<br> 9^1/2b/3^a

OlgaM077 [116]

Answer:

$\frac{1}{ 9 }$

Step-by-step explanation:

$\huge \frac{ {9}^{ \frac{1}{2}b } }{ {3}^{a} } \\ \\ = \huge \frac{ {( {3}^{2} )}^{ \frac{1}{2}b } }{ {3}^{a} } \\ \\ = \huge \frac{ { {3}}^{2 \times \frac{1}{2}b } }{ {3}^{a} } \\ \\ \huge = \frac{ {3}^{b} }{ {3}^{a} } \\ \\ \huge = \frac{1}{ {3}^{a - b} } \\ \\ \huge= \frac{1}{ {3}^{2} } \\ ( \because \: a - b = 2)\\ \\ \huge = \frac{1}{ 9 }$

5 0

3 years ago