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3 years ago

Which is larger 11.4cm or 118m?

Mathematics

1 answer:

VikaD [51]3 years ago

6 0

118m is larger than the 11.4cm as m is always greater than the CM

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Translate the statement into an equation or inequality. The sum of 5 and a number is less than the product of the number and 7 m

RideAnS [48]

Answer:

2x-4=10

2x=14

x=7

2. 3x+3=2x-1

3x-2x=-1-3

x=-4

3. 7x+x=24

8x=24

x=3(

4. 5+x=-18

x=-18-5

x=-23

5. -14=10-6x

6x=10+14

6x=24

x=4

6. 2x-2=x+12

2x-x=12+2

x=14

7. 3x-31=2

3x=2+31

3x=33

x=11

8. 5x-14=16

5x=16+14

5x=30

x=6

9. 2x+8=4(5-x)

2x+8=20-4x

2x+4x=20-8

6x=12

x=2

10. 2x-3=3(1+x)

2x-3=3+3x

2x-3x=3+3

-x=6

4 0

3 years ago

What is the difference? StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction Sta

katovenus [111]

Answer:

The option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is $\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

$\frac{x}{x^2-16}-\frac{3}{x-4}$

To solve the given expression

$\frac{x}{x^2-16}-\frac{3}{x-4}$

$=\frac{x}{x^2-4^2}-\frac{3}{x-4}$

$=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}$ ( using the property $a^2-b^2=(a+b)(a-b)$ )

$=\frac{x-3(x+4)}{(x+4)(x-4)}$

$=\frac{x-3x-12}{(x+4)(x-4)}$ ( by using distributive property )

$=\frac{-2x-12}{(x+4)(x-4)}$

$=\frac{-2(x+6)}{(x+4)(x-4)}$

$\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is $\frac{-2(x+6)}{(x+4)(x-4)}$

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4 years ago

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