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3 years ago

What is the solution to the linear equation 4b + 6 = 2 - b+ 4

Mathematics

1 answer:

inessss [21]3 years ago

5 0

Final Answer: $b = 0$

Steps/Reasons/Explanation:

Question: What is the solution to the linear equation $4b + 6 = 2 - b + 4$ ?

Step 1: Simplify $2 - b + 4$ to $6 - b$ .

$4b + 6 = 6 - b$

Step 2: Cancel $6$ on both sides.

$4b = -b$

Step 3: Subtract $-b$ from both sides.

$4b + b = 0$

Step 4: Simplify $4b + b$ to $5b$ .

$5b = 0$

Step 5: Divide both sides by $5$ .

$b = 0$

~I hope I helped you :)~

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Please help me with this math problem. I have to Use the quadratic formula to solve each equation.

Ierofanga [76]

You have to solve the given expression for q:

$2q^2-8=3q$

This expression has a quadratic term, which means that it is a quadratic equation. To find the value or values of q, you have to use the quadratic equation, using "q" as the variable instead of "x"

$q=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

a is the coefficient of the quadratic term

b is the coefficient of the q term

c is the constant

- First, zero the equation by passing 3q to the left side of the equal sign:

$\begin{gathered} 2q^2-8=3q \\ 2q^2-8-3q=3q-3q \\ 2q^2-3q-8=0 \end{gathered}$

For this equation the coefficients are:

a= 2

b= -3

c= -8

Replace them in the formula and solve:

$\begin{gathered} q=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ q=\frac{-(-3)\pm\sqrt{(-3)^2-4*2*(-8)}}{2*2} \\ q=\frac{3\pm\sqrt{9+64}}{4} \\ q=\frac{3\pm\sqrt{73}}{4} \end{gathered}$

Next, to determine each possible value of q, you have to solve the sum and difference separately:

Sum

$\begin{gathered} q=\frac{3+\sqrt{73}}{4} \\ q=2.886\cong2.9 \end{gathered}$

Difference

$\begin{gathered} q=\frac{3-\sqrt{73}}{4} \\ q=-1.386\cong-1.4 \end{gathered}$

The possible solutions for the given equation are q=2.9 and q=-1.4

4 0

1 year ago

Dy/dx y = 3root(4x+1) -2x

scoray [572]

Answer:

Step-by-step explanation:

y = 3(4x + 1)^1/2 - 2x

dy/dx = 3*1/2(4x + 1)^-1/2 * 4 - 2

= 4 * 1.5 (4x + 1)^-1/2 - 2

= 6(4x + 1)^-1/2 - 2

= [6/√(4x + 1)] - 2

y = 3(4x + 1)^1/2 - 2x

∫3(4x + 1)^1/2 - 2x . dx

= 3 * 1/4 * (4x + 1)^3/2 / 3/2 - x^2 + C

= 3/4 * 2/3 (4x + 1)^3/2 - x^2 + C

= 1/2((4x + 1)^3/2 - 2x^2) + C.

5 0

3 years ago

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denpristay [2]

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