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Mashutka [201]
3 years ago
11

Can you help me with this its easy

Mathematics
2 answers:
Darina [25.2K]3 years ago
4 0

Answer:

C. 1/8

Step-by-step explanation:

You gotta find the LCD and combine

Stolb23 [73]3 years ago
3 0

Answer:

There is no file attached

Step-by-step explanation:

I can't help if there's no file.

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Find c. Round to the nearest tenth.
mixer [17]

Answer:

Here,

150+10=180

162-180=18

Answer is 18°

Step-by-step explanation:

So the round of nearest tenth is 18°

Hope it will help have a great day at school. ^_^

3 0
3 years ago
Three more than the quotient of a number and 6 is 8.
meriva

Answer:

x/6 + 3 = 8

x/6 = 5

x = 30

Step-by-step explanation:

6 0
3 years ago
Find the least common multiple of 4 and 20
Svetach [21]

Answer:

Heya mate....

Step-by-step explanation:

This is ur answer.....

<h2>LCM of 4 and 20 is 20.</h2>

Hope it helps!

Brainliest pls!

Follow me :)

3 0
2 years ago
Read 2 more answers
Which is a reasonable first step that can be used to solve the equation 2(x + 6) = 3(x - 4) + 5?
snow_lady [41]

Answer:

Expose both of left side and right side.

For example: on the left side: 2(x+6) = 2x + 6.

6 0
3 years ago
Which equations have infinitely many solutions?
igomit [66]

Answer:

4(x - 1) = 4x - 4

3x + 6 = 3(x + 2)

Step-by-step explanation:

The first equation is

5x - 1 = 5x + 1

We simplify to get;

- 1 = 1

This is not true, therefore this equation has no solution.

The second equation is

3x - 1 = 1 - 3x

Combine like terms:

3x + 3x = 1 + 1

6x = 2

x =  \frac{1}{3}

This has a unique solution.

The 3rd equation is

7x - 2 = 2x - 7

Group similar terms:

7x - 2x =  7 + 2 \\ 5x = 9 \\ x  =  \frac{9}{5}

The 4th equation is :

4(x - 1) = 4x - 4

4x - 4= 4x - 4 \\ x = x

This is always true. The equation has infinite solution.

The 5th equation is:

3x + 6 = 3(x + 2) \\ 3x + 6 = 3x + 6 \\ x = x

This also has infinite solution

The 6th equation is

3(x - 4) = 4(x - 3) \\ 3x - 12 = 4x - 12 \\ 3x - 4x =  - 12 + 12 \\ x = 0

It has a unique solution.

8 0
3 years ago
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