Solve the following equation for x 10=2-4(ax-3)

Mathematics

1 answer:

likoan [24]2 years ago

7 0

Answer:

$\large\boxed{x=\dfrac{1}{a}}$

Step-by-step explanation:

$2-4(ax-3)=10\qquad\text{subtract 2 from both sides}\\\\-4(ax-3)=8\qquad\text{divide both sides by (-4)}\\\\\dfrac{-4(ax-3)}{-4}=\dfrac{8}{-4}\\\\ax-3=-2\qquad\text{add 3 to both sides}\\\\ax-3+3=-2+3\\\\ax=1\qquad\text{divide both sides by}\ a\neq0\\\\x=\dfrac{1}{a}$

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Sophie practices the piano 105 minutes in 1 week. If t represents the total time she practices for any number of days, d, write

Alex Ar [27]

Answer:

Step-by-step explanation:

You must first know that 1 week = 7days

If Sophie practices the piano 105 minutes in 1 week, this can be expressed as;

105 minutes = 7days

to calculate the total time t that she practices for any number of days, d, we will say;

t minutes = d days.

Solving both expression;

105 minutes = 7days

t minutes = d days.

Cross multiply;

7 * t = 105 * d

7t = 105d

Divide both sides by 7;

7t/7 = 105d/7

t = 15d

Hence a proportional equation for t in terms of d that matches the context is expressed as t = 15d

3 0

3 years ago

! + 3 − 5 = 0; by completing the square

mylen [45]

Answer:

x = 1.19258 and x = −4.19258

Step-by-step explanation:

The given equation is :

$x^2+3x-5=0$

Keep x terms on the left and move the constant to the right side by adding it on both sides.

$x^2+3x=5$

Take half of the x term and square it.

$(3\times \dfrac{1}{2})^2=\dfrac{9}{4}$

then add the result to both sides

$x^2+3x+\dfrac{9}{4}=5+\dfrac{9}{4}\\\\(x+\dfrac{3}{2})^2=\dfrac{29}{4}\\\\(x+\dfrac{3}{2})=\sqrt{\dfrac{29}{4}}\\\\x=\dfrac{-3}{2}+\sqrt{\dfrac{29}{4}}\\\\x=\dfrac{-3}{2}-\sqrt{\dfrac{29}{4}}$