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2 years ago

3(8k – 3) = -6(7 -- 4k)

Mathematics

2 answers:

elixir [45]2 years ago

4 0

I think you want me to solve for k

expanding, you get

24k-9=-42+24k

which is clearly no solution

Arada [10]2 years ago

4 0

I got that the statement is false for any value of k !!
hope that helps you out :D

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Tell whether the relation is a function (3,-2)(0,1)(1,0)(-2,-1)(2,-1)

GREYUIT [131]

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3 years ago

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3 years ago

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3 years ago

What is the solution to -x2 + 5x - 3

CaHeK987 [17]

The solution to the equation is $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Explanation:

Given that the equation is $-x^2+5x-3=0$

We need to determine the solution of the equation.

The solution of the equation can be determined using the quadratic formula.

The quadratic formula is given by

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

Hence, from the equation, we have,

$a=-1,\:b=5,\:c=-3$

Substituting these values in the quadratic formula, we get,

$x=\frac{-5\pm \sqrt{5^2-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$

Simplifying, we get,

$x=\frac{-5\pm \sqrt{25-12}}{-2}$

Simplifying the terms within the root, we get,

$x=\frac{-5\pm \sqrt{13}}{-2}$

Thus, the roots of the equation are $x=\frac{-5+ \sqrt{13}}{-2}$ and $x=\frac{-5- \sqrt{13}}{-2}$

Taking out the negative sign, we get,

$x=\frac{-(5- \sqrt{13})}{-2}$ and $x=\frac{-(5+ \sqrt{13})}{-2}$

Cancelling the negative sign, we get,

$x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Thus, the solutions of the equation are $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

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4 years ago

Solve -4b + 2 = 18.

ratelena [41]

B=-4 subtract 2 form both sides which gives u -4b=16 then divided both sides by -4 B=-4

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3 years ago

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