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3 years ago

Which is the solution to the equation 1/4x-1/8=7/8+1/2x

Mathematics

2 answers:

ivann1987 [24]3 years ago

5 0

Answer:

x = -4

Step-by-step explanation:

1/4x-1/8=7/8+1/2x

Add 1/8 to each side

1/4x-1/8+1/8=7/8+1/8+1/2x

1/4 x = 1 + 1/2x

Subtract 1/2x from each side

1/4x -1/2x = 1 + 1/2x -1/2x

Getting a common denominator for the x terms

1/4x -2/4x =1

-1/4x = 1

Multiply by -4

-4 * -1/4x = 1* -4

x = -4

timofeeve [1]3 years ago

4 0

Answer:

x = -4

Step-by-step explanation:

Step 1: Subtract 1/2x on both sides

-1/4x - 1/8 = 7/8

Step 2: Add 1/8 on both sides

-1/4x = 1

Step 3: Divide both sides by -1/4

x = -4

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timurjin [86]

Answer:

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Step-by-step explanation:

-10+20=10

10+-40=-30

-30+80=50

5 0

3 years ago

Select ALL the correct answers.

IgorC [24]

Answer:

A. 4(x - 4) + 2(3x² + 3x - 20)

C. (11x² + 7x - 55)-(5x² - 3x + 1)

F. (3x² + 5x - 28) + (3x² + 5x - 28)

Step-by-step explanation:

Given:

(3x-7)(2x+8)

= 6x² + 24x - 14x - 56

=6x² + 10x - 56

A. 4(x - 4) + 2(3x² + 3x - 20)

= 4x - 16 + 6x² + 6x - 40

= 6x² + 10x - 56

B. (3x² + 5x - 28) - (2x² + 4x + 28)

= 3x² + 5x - 28 - 2x² - 4x - 28

= x² + x - 56

C. (11x² + 7x - 55)-(5x² - 3x + 1)

= 11x² + 7x - 55 - 5x² + 3x - 1

= 6x² + 10x - 56

D. 4(x - 4) - 2(3x² + 3x - 20)

= 4x - 16 - 6x² - 6x + 40

= - 6x² - 2x + 24

E. (11x² + 7x - 55)-(5x² - 3x + 2)

= 11x² + 7x - 55 - 5x² + 3x - 2

= 11x² - 5x² + 7x + 3x - 55 - 2

= 6x² + 10x - 57

F. (3x² + 5x - 28) + (3x² + 5x - 28)

= 3x² + 5x - 28 + 3x² + 5x - 28

= 6x² + 10x - 56

7 0

3 years ago

Determine if the lim f(x) exists using the graph below.. If it does, find its value. If it does not,x 1explain why

Leno4ka [110]

The limit of a function f at a point a exists if:

$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)$ $\begin{gathered} \text{ From the given image, the value of the left limit of f at x=1 } \\ \lim_{x\to a^-}f(x)\lt1 \end{gathered}$

Also,

$\begin{gathered} \text{ From the given image, the value of the right limit of f at x=1 } \\ \lim_{x\to a^+}f(x)\gt1 \end{gathered}$

Therefore,

$\lim_{x\to1^-}f(x)\lim_{x\to1^+}f(x)$

Hence, the limit does not exist

8 0

1 year ago

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Lorico [155]

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Step-by-step explanation:

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3 years ago

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tresset_1 [31]

4^12 = 16777216 is the answer.

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