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2 years ago

11

How do I solve this x/2-1=x/5+2

1 answer:

Ainat [17]2 years ago

6 0

To solve this equation you need to solve for the variable x. There are more than one way you can to this but I find substitution method to be viable.

x=10, we know this because 10/2= 5 and 5-1=4. Also 10/5=2 and 2+2=4.

please vote my answer brainliest. thanks!

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Alicia buys a 5- pounds bag of rocks for a fish tank. She uses 1 1/8 pounds for a small fish bowl. How much is left?

Alex73 [517]

5 subtract 1 and one eighth is 3 and seven eighths
3 7/8

7 0

2 years ago

Please solve this fast late night homework session and I need to get as much done as possible...

aksik [14]

Answer:

-4

Step-by-step explanation:

4x+3y=-1

3x + y=-7

Solve for Y

3x+y=-7

-3x. -3x

y= -3x - 7

Plug in Y into the top equation

4x + 3 (-3x - 7) = -1

Distribute the Parenthesis

4x -9x-21=-1

Combine Like terms

-5x = 20

Divide by -5 on both sides

20/-5

x = -4

7 0

2 years ago

Sophia is organizing textbooks on her bookshelf. She has an English textbook, a biology textbook, a history textbook, and a writ

ANTONII [103]

Answer:

I believe it's 24 combinations.

Step-by-step explanation:

I found how many there were when starting with English, and there were six. I multiplied this by four (for the four books) and found 24.

Hope this helps! Have a nice day :)

5 0

2 years ago

Read 2 more answers

Can someone help me answer this?

mart [117]

Answer:

x = 40 , y = 70

Step-by-step explanation:

y = 70°

x° = 180° - 2 × 70° = 40°

5 0

3 years ago

Read 2 more answers

At what values of x does f(x) = x^3 - 2x^2 -4x+1 satisfy the mean value theorwm on [0,1]

denis-greek [22]

Answer:

x=1/3

Step-by-step explanation:

A function f is given as

$f(x) = x^3-2x^2-4x+1$ in the interval [0,1]

This function f being an algebraic polynomial is continuous in the interval [0,1] and also f is differntiable in the open interval (0,1)

Hence mean value theorem applies for f in the given interval

$f(1) = 1-2-4+1 = -4\\f(0) = 1$

The value

$\frac{f(1)-f(0)}{1-0} =\frac{-4-1}{1} =-5$

Find derivative for f

$f'(x) = 3x^2-4x-4$

Equate this to -5 to check mean value theorem

$3x^2-4x-4=-5\\3x^2-4x+1=0\\\\(x-1)(3x-1) =0\\x= 1/3 : x = 1$

We find that 1/3 lies inside the interval (0,1)

4 0

2 years ago