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3 years ago

How many solutions exist for the given equation 1/2 (x+12)=4x-1

Mathematics

2 answers:

Daniel [21]3 years ago

7 0

Answer:

x=2

Step-by-step explanation:

1/2 (x+12)=4x-1

Multiply each side by 2

2*1/2 (x+12)=2*(4x-1)

x+12 = 8x -2

x+12 = 8x - 2

Subtract x from each side

12 = 7x -2

Add 2 to each side

14 =7x

Divide by 7

14/7 = 7x/7

2 =x

IRISSAK [1]3 years ago

7 0

Steps to solve:

1/2(x + 12) = 4x - 1

~Distribute

1/2x + 6 = 4x - 1

~Subtract 6 to both sides

1/2x = 4x - 7

~Subtract 4x to both sides

-7/2x = -7

~Multiply -2/7 to both sides

x = 2

Best of Luck!

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In standard form.3+0.4+0.05+0.002 has which digit in the hundredths place

klemol [59]

Answer:

Step-by-step explanation:

0.4

0.05

0.002= 3.452

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4 years ago

Which numbers are solutions to the inequality -3x-3<6? Check all that apply

bogdanovich [222]

Answer:

0, 2, and 4.

Step-by-step explanation:

-3x-3≤6

Add 3 on both sides of the inequality.

-3x≤9

Divide both sides of the inequality by -3.

x≥3

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3 years ago

X + y = 3 2x + 2y = -4

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A specific brand of gourmet chocolate candy contains 7.00 gg of dietary fat in each 22.7-gg piece. How many kilograms of dietary

Oduvanchick [21]

Answer:

635599482kg

Step-by-step explanation:

A gourmet chocolate candy has 7.00 gg of dietary fat in each 22.7-gg piece

for the 22.7 piece , the gg is 7 x 22.7 = 158.9gg in total

1gg = 2204622.6218488 lbs

158.9gg in total = 158.9 x 2204622.6218488 = 350314534.612

for the box of 4.00 lblb of candy

it is 350314534.612 x 4 = 1401258138.45lb

since our answer is needed in kilograms we convert 1401258138.45lb to kilograms

1lb = 0.453592kg

1401258138.45lb = 1401258138.45lb x 0.453592kg = 635599481.513 ≈ 635599482kg

5 0

4 years ago

If G^-1(x) is the inverse of G(x), what is the value of G^-1(G(x))?

Ber [7]

Answer:

$G^{-1}(G(x))=x$

Step-by-step explanation:

Definition: Let f be one-to-one function with the domain A and the range B. Then the inverse function $f^{-1}$ has the domain B and the range A and is defined as

$f^{-1}(y)=x\ \Leftrightarrow \ f(x)=y$

Facts:

1) For every $x\in A:$

$f^{-1}(f(x))=x$

2) For every $y\in B:$

$f(f^{-1}(y))=y$

So, $G^{-1}(G(x))=x$

5 0

4 years ago