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Marizza181 [45]
3 years ago
15

(−5)3x7(yz)4 / (3)2x2y8z2

Mathematics
1 answer:
g100num [7]3 years ago
5 0

Answer:

\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}

Step-by-step explanation:

Given:

The expression to simplify is given as:

\frac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}

In order to simplify this, we have to use the law of indices.

1. (ab)^m=a^mb^m

So, (yz)^4=y^4z^4

Substitute this value in the above expression. This gives,

=\dfrac{(-5)3x^7y^4z^4}{(3)2x^2y^8z^2}\\\\\\=\dfrac{-15x^7y^4z^4}{6x^2y^8z^2}......(-5\times 3=15\ and\ 3\times 2=6)

Now, we use another law of indices.

2. \frac{a^m}{a^n}=a^{m-n}

So,  \frac{x^7}{x^2}=x^{7-2}=x^5,\frac{y^4}{y^8}=y^{4-8}=y^{-4}, \frac{z^4}{z^2}=z^{4-2}=z^2

Substitute these values in the above expression. This gives,

=\frac{-15}{6}\times x^5\times y^{-4}\times z^2\\\\=\frac{-5x^5y^{-4}z^2}{2}

Finally, we further simplify it using the law a^{-m}=\frac{1}{a^m}

So, y^{-4}=\frac{1}{y^4}

Therefore, the given expression is simplified as:

\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}

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Determine the gradient of the straight line 2x-3y+9=0. Find the equation of the straight line through the origin which is perpen
frutty [35]

Answer: 3

x

−

2

y

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15

=

0

Explanation:

We know that,

the slope of the line  

a

x

+

b

y

+

c

=

0

is  

m

=

−

a

b

∴

The slope of the line  

2

x

+

3

y

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9

is  

m

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=

−

2

3

∴

The slope of the line perpendicular to  

2

x

+

3

y

=

9

is  

m

2

=

−

1

m

1

=

−

1

−

2

3

=

3

2

.

Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

with slope

m

is

y

−

y

1

=

m

(

x

−

x

1

)3

x

−

2

y

−

15

=

0

Explanation:

We know that,

the slope of the line  

a

x

+

b

y

+

c

=

0

is  

m

=

−

a

b

∴

The slope of the line  

2

x

+

3

y

=

9

is  

m

1

=

−

2

3

∴

The slope of the line perpendicular to  

2

x

+

3

y

=

9

is  

m

2

=

−

1

m

1

=

−

1

−

2

3

=

3

2

.

Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

with slope

m

is

y

−

y

1

=

m

(

x

−

Explanation:

the equation of a line in  

slope-intercept form

is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

−

2

3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

perpendicular

=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular lineExplanation:

the equation of a line in  

slope-intercept form

is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

−

2

3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

perpendicular

=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular line

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