Solve for the value of y.

With full explanation from the internet like before 3x2-6x+5=0

Elodia [21]

Answer:

$\sf x=1+i\sqrt{\dfrac{2}{3}} \ \quad and \quad \:x=1-i\sqrt{\dfrac{2}{3}}$

Explanation:

Given Expression:

3x² - 6x + 5 = 0

Use the Quadratic Formula:

$\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \ \ when \ \ ax^2 + bx + c = 0$

insert coefficients

$\Longrightarrow \sf x = \dfrac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}$

$\Longrightarrow \sf x = \dfrac{\left6\right\pm \sqrt{-24} }{6}$

$\Longrightarrow \sf x = \dfrac{\left6\right\pm 2\sqrt{6}i}{6}$

$\Longrightarrow \sf x =1 \pm i\dfrac{\sqrt{6} }{3}$

$\Longrightarrow \sf x=1+i\sqrt{\dfrac{2}{3}}, \quad 1-i\sqrt{\dfrac{2}{3}}$

6 0

2 years ago

Read 2 more answers

Please help with this and explain how to do it, I'll give 15 points

MrRissso [65]

Answer:

Option (4).

Step-by-step explanation:

Outer diameter of the hollow metallic ball = 10 centimeters

Outer radius of this ball = $\frac{10}{2}$ = 5 cm

Volume of the outer ball $V_{1}$ = $\frac{4}{3}\pi (r)^3$

= $\frac{4}{3}\pi (5)^3$

Inner radius of the hollow metallic ball = (5 - 1) = 4 cm

Volume of the inner hollow ball $V_{2}$ = $\frac{4}{3}\pi (4)^3$

Volume of the metal used in the metallic ball = $V_{1}-V_{2}$

= $\frac{4}{3}\pi (5)^3-\frac{4}{3}\pi (4)^{3}$

= $\frac{4}{3}\pi [(5)^3-(4)^3]$

Therefore, expression given in option (4) will be used to measure the volume of the hollow metallic ball.

8 0

3 years ago

60 billetes de a 100 cuantos miles son

vladimir1956 [14]

Si multiplicas 60 por 100 obtienes 6000 por lo que hay 6 mil

7 0

3 years ago

Read 2 more answers

Draw a model to represent the problem

Mazyrski [523]

2 because if you convert the fractions into decimals then divide it that way it will be 2

6 0

4 years ago

Help please!! i don’t understand this one!

12345 [234]

Answer:

C, D

Step-by-step explanation:

This equation can be solved any of several ways. One that doesn't require much thought is using the quadratic formula.

For ax² +bx +c = 0, the solutions are ...

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

In the given equation, a=2, b=11, c=5, so this becomes ...

$x=\dfrac{-11\pm\sqrt{11^2-4\cdot2\cdot5}}{2\cdot2}=\dfrac{-11\pm\sqrt{81}}{4}\\\\x=\dfrac{-11\pm9}{4}=\left\{-\dfrac{20}{4},-\dfrac{2}{4}\right\}=\{-5,-\frac{1}{2}\}$

The solutions are ...

C. x = -5

D. x = -1/2

_____

My personal favorite is using a graphing calculator. The solutions are the x-intercepts of the expression on the left. That is, where its value is zero, as the equation says.

5 0

3 years ago