\frac{(2_1-4_{2}) ^{4} } {2} Quick I need help Plz answer it NOW

Mathematics

1 answer:

erastova [34]3 years ago

6 0

Answer:

2(2^1-4^2)^4

Step-by-step explanation:

You might be interested in

Which of the following statements accurately describe the dilation shown? Select all that apply.

Anon25 [30]

Answer:

A, D, and F I think. Hope this helps

8 0

2 years ago

Use the discriminant to determine what type of roots the equations will have, and categorize the equations according to their ro

topjm [15]

Step-by-step explanation:

The discriminant of the quadratic equation $ax^2+bx+c=0$ :

$\Delta=b^2-4ac$

If Δ < 0, then the equation has two complex roots $x=\dfrac{-b\pm\sqrt\Delta}{2a}$

If Δ = 0, then the equation has one repeated root $x=\dfrac{-b}{2a}[/tex If Δ > 0, then the equation has two discint roots [tex]x=\dfrac{-b\pm\sqrt\Delta}{2a}$

$1.\ x^2-4x+2=0\\\\a=1,\ b=-4,\ c=2\\\\\Delta=(-4)^2-4(1)(2)=16-8=8>0,\ \bold{two\ distinct\ roots}\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=2\sqrt2\\\\x=\dfrac{-(-4)\pm2\sqrt2}{2(1)}=\dfrac{4\pm2\sqrt2}{2}=2\pm\sqrt2\\\\==============================\\\\2.\ 5x^2-2x+3=0\\\\a=5,\ b=-2,\ c=3\\\\\Delta=(-2)^2-4(5)(3)=4-60=-56$

$3.\ 2x^2+x-6=0\\\\a=2,\ b=1,\ c=-6\\\\\Delta=1^2-4(2)(-6)=1+48=49>0,\ \bold{two\ distinct\ roots}\\\sqrt\Delta=\sqrt{49}=7\\\\x=\dfrac{-1\pm7}{(2)(2)}\\\\x_1=\dfrac{-8}{4}=-2,\ x_2=\dfrac{6}{4}=\dfrac{3}{2}\\\\==============================\\\\4.\ 13x^2-4=0\qquad\text{add 4 to both sides}\\\\13x^2=4\qquad\text{divide both sides by 13}\\\\x^2=\dfrac{4}{13}\to x=\pm\sqrt{\dfrac{4}{13}},\ \bold{two\ distinct\ roots}\\\\==============================$

$5.\ x^2-6x+16=0\\\\a=1,\ b=-6,\ c=16\\\\\Delta=(-6)^2-4(1)(16)=36-64=-28$

$7.\ 4x^2+11=0\qquad\text{subtract 11 from both sides}\\\\4x^2=-11\qquad\text{divide both sides by 4}\\\\x^2=-\dfrac{11}{4}\to x=\pm\sqrt{-\dfrac{11}{4}}\\\\x=\pm\dfrac{\sqrt{11}}{2}\ i,\ \bold{two\ complex\ roots}$