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1

     2 (x^2-2x) -6   complete the square

Y=  2 (x-1)^2 - 8

2

  x intercept(s) happen when y = 0   set the equation to y=0

  0 = 2(x-1)^2 - 8

4 = (x-1)^2

+- 2 = x-1

x = 3, -1

3

  From 2 above    (-inf,-1) U (3, +inf)

Range = all values of x are allowed in the domain...the range extends from the vertex at -8 to + inf     [-8, +inf)

This is an upright parabola....the funtion is decreasing from -inf to the vertex (1,-8)    or for values of x  ( -inf ,1]

h**l yeah bro

Note that BC = AC = CD

So......in triangle ACD, angle ACD = angle ACB + angle BCD=  60° + 90°  = 150°

And since  AC = CD.....then the angles opposite these sides in triangle ACD are also equal

So....angle  CAD = angle CDA

So....

angle CAD + angle CDA + angle ACD = 180

angle CAD + angle CAD + angle ACD =  180

2angle CAD + 150  = 180

2angle CAD = 180 - 150

2angle CAD = 30          divide both sides by 2

angle CAD = 15°

-4 -7m < 8 -5m        add 7m to both sides, subtract 8 from both sides

-12 < 2m            divide both sides by 2

-6 < m

Which is the same as

m > -6

16/n^2 * 2.352 = 425.32

Divide both sides by 2.352

16/n^2 = 425.32 / 2.352    and we can write

n^2 /16 = 2.352 / 425.32    multiply both sides by 16

n^2 =   16*2.352 / 425.32     take both roots 

n = ±√ [ 16* 2.352 / 425.32]  ≈   ± 0.297

Here is the graph :  https://www.wolframalpha.com/input/?i=16%2Fn%5E2%C2%A0*+2.352+%3D+425.32

14a^3- 3/a^2=426     [ I'll use "x" instead of "a" ]

14^x^3 -3/x^2 - 426 = 0      multiply through by x^2

14x^5 - 3 - 426x^2 = 0

14x^5 - 426x^2 - 3 = 0

This does not have an algebraic solution, DP   [ or, at least, I don't think that it does ]

There is only one real root at x = a  ≈ 3.12

There are also   4 complex roots

Here is the graph and solution  found by WolframAlpha :

https://www.wolframalpha.com/input/?i=14x%5E5+-+3+-+426x%5E2+%3D+0

Sorry, but I need a detailed explanation to this ASAP.

0.9 * 130 is 117.

117

The answer would be 14.25 dollars.

13.32 * 0.07 = 0.9324 (0.93 rounded to the nearest cent)

13.32 + 0.93 = 14.25

Rip.

I am a senior. Calculus is really hard. :/

The sky is up. r if your inside a building, probably a roof of some sort.

#1 - Solve for x:
0.0075 (x^3 - x^2) = 0

0.0075 (x^3 - x^2) = 3/400 (x^3 - x^2):
3/400 (x^3 - x^2) = 0

Multiply both sides by 400/3:
x^3 - x^2 = 0

Factor x^2 from the left hand side:
x^2 (x - 1) = 0

Split into two equations:
x - 1 = 0 or x^2 = 0

Add 1 to both sides:
x = 1 or x^2 = 0

Take the square root of both sides:

x = 1    or    x = 0

#2 - Solve for x:
0.0525 x^2 = 0

0.0525 x^2 = (21 x^2)/400:
(21 x^2)/400 = 0

Multiply both sides by 400/21:
x^2 = 0

Take the square root of both sides:

x = 0

We  have the form

y = a(x - h)^2 + k

Where   (h, k) = the vertex = ( -1, 6).....and we know a point on the graph... (0, -1)

So....we can solve for a

-1 = a(-1 - (0))^2  + 6

-1 = a( -1)^2 + 6

-1 = a + 6      subtract 6 from both sides

-7 = a

So....in vertex form, we have

y = - 7( x - (-1) )^2 + 6

y = -7 ( x + 1)^2 + 6

Hello! 

Area....

We have a quadrilateral with two parallel sides AD and BC...so....this is a trapezoid

The area is

(1/2) height of trapezoid * ( sum of the base lengths)

The bases are DC and AB.....we know these

The height is the tricky part....it is found as:

[ The y coordinate of B - The y coordinate of C ] =  [  1 - (-5) ] = [ 1 + 5] = 6 units

So.....the area is

(1/2) (6) ( 3 + 2√5 )   ≈    22.42 units^2

And that's it....!!!

Same here

One of my favorite tunes......!!!!!

AD is correct !!!

And you got BC correct.... we can simplify it as   3√5

Let's look at DC.....

√ [ ( -4 - (-1) )^2 + ( -5 - (-5) )^2 ] =  √ [ ( -3)^2 + ( 0 )^2] = √9 = 3

So.... the perimeter is

AB + AD + DC + BC =

2√5 + 2√5 + 3 + 3√5 =

7√5 + 3   units ≈  18.65 units

You did pretty well....!!!!

The area is a little tricky, RP.....I'll get back to it in just a bit.....I have a few more questions to get to....just hold on...!!!

I hear the drums echoing tonight
But she hears only whispers of some quiet conversation
She's coming in, 12:30 flight
The moonlit wings reflect the stars that guide me towards salvation
I stopped an old man along the way
Hoping to find some long forgotten words or ancient melodies
He turned to me as if to say, "Hurry boy, it's waiting there for you"

It's gonna take a lot to take me away from you
There's nothing that a hundred men or more could ever do
I bless the rains down in Africa
Gonna take some time to do the things we never had

The wild dogs cry out in the night
As they grow restless, longing for some solitary company
I know that I must do what's right
As sure as Kilimanjaro rises like Olympus above the Serengeti
I seek to cure what's deep inside, frightened of this thing that I've become

It's gonna take a lot to drag me away from you
There's nothing that a hundred men or more could ever do
I bless the rains down in Africa
Gonna take some time to do the things we never had

Hurry boy, she's waiting there for you

It's gonna take a lot to drag me away from you
There's nothing that a hundred men or more could ever do
I bless the rains down in Africa
I bless the rains down in Africa
(I bless the rain)
I bless the rains down in Africa
(I bless the rain)
I bless the rains down in Africa
I bless the rains down in Africa
(Ah, gonna take the time)
Gonna take some time to do the things we never had

We'll try, Toto.....!!!

Since SR parallel to PQ....they are equal

So....to find PS = QR we have

24 - 2SR = PS + QR

24 -2(5) = PS + QR

24 - 10 = PS + QR

14 = PS + QR

So  QR = 7 = PS

And the area of the parallelogram PQRS =  QR * ST

So

28 = 7 * ST    divide both sides by 7

4 = ST = OQ

And using the Pythagorean Theorem

RT =  sqrt [ RS^2 - ST^2] =  sqrt [ 5^2 - 4^2] = sqrt [ 25 - 16] = sqrt[9] = 3 = OP

So....the perimeter of QOST =

2 (OP + PS) + 2(ST) =

2 (3 + 7) + 2(4) =

2*10 + 2*4 =

28 units

Go commit no breathe

I got:

2sqrt(5) for A and D

sqrt(3i) for D and C

C and B is sqrt(9) sqrt(5)