Please help quick! Write the quadratic function f(x) = 3(x + 1)(x – 7) in standard form. (Answers attached below)

Zielflug [23.3K]

The quadratic formula is
$x= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$ .

For the equation $4x^2-3x+9=2x+1$ , we must first convert it into standard form $ax^2+bx+c=0$ . We see it is $4x^2-5x+8=0$ . From this, our coefficients are $a=4, b=-5$ and $c=8$ .

We plug these coefficients into the quadratic formula.

$x= \dfrac{-(-5)\pm\sqrt{(-5)^2-4(4)(8)} }{2(4)} \\ \\ \\ \\ x = \dfrac{5\pm\sqrt{-103}}{8} \\ \\ \\ \\ x = \dfrac{5\pm\ i\sqrt{103}}{8}$

The answer is C.

3 0

3 years ago

Calculate the slope of the line between each pair of points (1,10) and (2,6)

skelet666 [1.2K]

Answer:

Remember those walls I built

Well, baby, they're tumbling down

And they didn't even put up a fight

They didn't even make a sound

I found a way to let you win

But I never really had a doubt

Standing in the light of your halo

I got my angel now

It's like I've been awakened

Every rule I had you breaking

It's the risk that I'm taking

I ain't never gonna shut you out

Everywhere I'm looking now

I'm surrounded by your embrace

Baby, I can see your halo

You know you're my saving grace

You're everything I need and more

It's written all over your face

Baby, I can feel your halo

Pray it won't fade away

I can feel your halo (halo) halo

I can see your halo (halo) halo

I can feel your halo (halo) halo

I can see your halo (halo) halo

Hit me like a ray of sun

Burning through my darkest night

You're the only one that I want

Think I'm addicted to your light

I swore I'd never fall again

But this don't even feel like falling

Gravity can't forget

To pull me back to the ground again

Feels like I've been awakened

Every rule I had you breaking

The risk that I'm taking

I'm never gonna shut you out

Everywhere I'm looking now

I'm surrounded by your embrace

Baby, I can see your halo

You know you're my saving grace

You're everything I need and more

It's written all over your face

Baby, I can feel your halo

Pray it won't fade away

I can feel your halo (halo) halo

I can see your halo (halo) halo

I can feel your halo (halo) halo

I can see your halo (halo) halo

I can feel your halo (halo) halo

I can see your halo (halo) halo

I can feel your halo (halo) halo

I can see your halo (halo) halo

Halo, halo

Ooh, ooh, ooh, ooh, ooh

Everywhere I'm looking now

I'm surrounded by your embrace

Baby, I can see your halo

You know you're my saving grace

You're everything I need and more

It's written all over your face

Baby, I can feel your halo

Pray it won't fade away

I can feel your halo (halo) halo

I can see your halo (halo) halo

I can feel your halo (halo) halo

I can see your halo (halo) halo

I can feel your halo (halo) halo

I can see your halo (halo) halo

I can feel your halo (halo) halo

I can see your halo (halo) halo

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Please help with sum plz

belka [17]

The angle is the same

all triangles equal to 180 degrees, so logically, the other angle is 60

It is an equilateral triangle.

Please add me as the brainliest!

Dey are the same exact triangle

3 0

2 years ago

Jessica borrow $316 for 9 months he paid it off by paying $38.27 monthly what will she pay in finance charges

liraira [26]

Answer:

my answer is 34 r 18 I am guessing

6 0

2 years ago

Read 2 more answers

A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t

sladkih [1.3K]

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

a)

$\frac{dy}{dt}\ \alpha\ y(1-y)$

$\frac{dy}{dt}=ky(1-y)$

where k is the constant of proportionality, dy/dt = rate at which the rumor spreads

b)

$\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\$

$y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}$

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

$y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days$

5 0

2 years ago