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3 years ago

What is the solution to the liner equation -12+3b-1=-5-b

Mathematics

2 answers:

Natalija [7]3 years ago

6 0

Answer:

b=2

Step-by-step explanation:

-12+3b-1=-5-b

Combine like terms

-13 +3b = -5-b

Add b to each side

-13+3b+b = -5-b+b

-13+4b = -5

Add 13 to each side

-13+13 +4b = -5+13

4b = 8

Divide by 4

4b/4 = 8/4

b = 2

olchik [2.2K]3 years ago

5 0

Answer: B=2

Step-by-step explanation:

-12+3b-1=-5-b Equation

-5+12+1=8 Adding the common number together

3b+b=4b Adding the b's together

4b=8 solution

/4 to each side (division)

b=2

When you add the numbers together it adds or subtracts from both sides so it will end up being the opposite of what it was originally ( all negative become positive. But will be on the other side)

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DedPeter [7]

Answer:

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Step-by-step explanation:

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3 years ago

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Andrej [43]

Sorry but this isnt even a full question.......

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3 years ago

Is P - 1 upon p = 4 find P + 1 upon p whole square

zlopas [31]

Given that (p - 1/p) = 4, the value of p² + 1/p² is 18. Detail below

<h3>Data obtained from the questio</h3>

(p - 1/p) = 4
p² + 1/p² = ?

<h3>How to determine the value of p² + 1/p²</h3>

(p - 1/p) = 4

Square both sides

(p - 1/p)² = (4)²

(p - 1/p)² = 16 ....(1)

Recall

(a - b)² = a² + b² - 2ab

Thus,

(p - 1/p)² = p² + 1/p² - (2 × p × 1/p)

(p - 1/p)² = p² + 1/p² - 2

From equation (1) above,

(p - 1/p)² = 16

Therefore,

p² + 1/p² - 2 = 16

Rearrange

p² + 1/p² = 16 + 2

p² + 1/p² = 18

Thus, the value of p² + 1/p² is 18

Learn more about algebra:

brainly.com/question/953809

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2 years ago

Which situation can be represented by Y= 12x - 4?

andrey2020 [161]

Answer:

D. The number of eggs , y, in x, dozen eggs for sale after 4 dozen eggs are sold

Step-by-step explanation:

The correct answer is D

The number of eggs , y, in x, dozen eggs for sale after 4 dozen eggs are sold.

Hope this helps.

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3 years ago

Read 2 more answers

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

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3 years ago