3 years ago

-ax+3b>5 solve for x

Mathematics

1 answer:

adoni [48]3 years ago

4 0

Answer:

x < 3b/a − 5/a

You might be interested in

Which of the following are solutions to the equation below?

svet-max [94.6K]

The answers are A and C

8 0

4 years ago

(23+8x-x²)+(x³+x²+34-18x)

bearhunter [10]

Answer:

x^3 -10x + 57

Step-by-step explanation:

3 0

3 years ago

What's the simplified form of (2x) -4

DIA [1.3K]

2x - 4

Hope it helps.

3 0

4 years ago

The Westhafen tower in Germany is in the shape of a cylinder. It has a height of 109 meters and a diameter of 38 meters. What is

aniked [119]

Answer:

123619 m^3

Step-by-step explanation:

V=π(d /2)2h=π·(38 /2)2·109 ≈ 1.23619×105m³

Not sure this is right

7 0

3 years ago

Read 2 more answers

Using the quadratic formula to solve 4x2 – 3x + 9 = 2x + 1, what are the values of x? StartRoot 1 plus-or-minus StartRoot 159 En

castortr0y [4]

For this case we have that a quadratic equation is of the form:

$ax ^ 2 + bx + c = 0$

The roots are given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

We have the following equation:

$4x ^ 2-3x + 9 = 2x + 1\\4x ^ 2-3x-2x + 9-1 = 0\\4x ^ 2-5x + 8 = 0$

We look for the roots:

$x = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (4) (8)}} {2 (4)}\\x = \frac {5 \pm \sqrt {25-128}} {8}\\x = \frac {5 \pm \sqrt {-103}} {8}$

We have to:

$i ^ 2 = -1$

So:

$x = \frac {5 \pm \sqrt {103i ^ 2}} {8}\\x = \frac {5 \pm i \sqrt {103}} {8}$

We have two imaginary roots:

$x_ {1} = \frac {5 \ + i \sqrt {103}} {8}\\x_ {2} = \frac {5 \ -i \sqrt {103}} {8}$

Answer:

$x_ {1} = \frac {5 \ + i \sqrt {103}} {8}\\x_ {2} = \frac {5 \ -i \sqrt {103}} {8}$

8 0

4 years ago

Read 2 more answers