All categories

2 years ago

2x-3x²=5 in standard formation

Mathematics

2 answers:

harina [27]2 years ago

5 0

Answer:

Step-by-step explanation:

1. Make the equation equal to zero

2x - 3x^2 = 5 = 0

2. Then apply simple algebra by subtracting -5 from +5

2x - 3x^2 = 5 = 0

-5

3. Now move the -5 to the left side of the equation

2x - 3x^2 -5 = 0

4. Now arrange the numbers into standard form (ax^2 + bx + c)

-3x^2 + 2x - 5

UkoKoshka [18]2 years ago

4 0

Bc 2x—x3=ew.r divorce

You might be interested in

Hi, Could anyone help me with my homework

tatiyna

Answer:

$\hookrightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

solving steps:

$\rightarrow \sf (x + 4)^6$

$\bold{rewrite \ the \ following}$

$\rightarrow \sf (x + 4)^2 (x + 4)^2 (x + 4)^2$

$\bold {formula \ used : \sf (x+a)^2 = (x^2 + 2xa + a^2)}$

$\rightarrow \sf (x^2 + 8x+16) (x^2 + 8x+16) (x^2 + 8x+16)$

$\bold{simplify \ by \ removing \ parenthesis}$

$\rightarrow \sf (x^4 +8x^3 + 16x^2 + 8x^3 +64x^2 + 128x+16x^2+128x+256 ) (x^2 + 8x+16)$

$\bold{basic \ addition \ of \ integers }$

$\rightarrow \sf (x^4+16x^3+96x^2+256x+256) (x^2 + 8x+16)$

$\bold{remove \ parenthesis}$

$\rightarrow \sf (x^6 + 16x^5 + 96x^4 + 256x^3 + 256x^2 + 8x^6 + 128x^4 + 768x^3 + 2048x^2 + 2048 + 16x^4 + 256x^3 + 1536x^2 + 4096x + 4096)$

$\bold {final \ answer:}$

$\rightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

8 0

2 years ago

How do you find the area of a traizoid

BaLLatris [955]

[(base + base)*high]/2

5 0

3 years ago

Someone help please I'm really stuck on these questions and I need help please I need to bring this in tomorrow please help

AysviL [449]

1) 1
2) PA: between 2 and 3.
2) PB: Unsure about it.

7 0

3 years ago

The volume of. Cube that is 23 in w,10 I’m D and 8 In H is

Masja [62]

23*10*8=1840
1840 cubic units

3 0

3 years ago

Read 2 more answers

The vertex of this parabola is at (-2, 5) (opens upward)

Rzqust [24]

Answer:

b. y = 3(x+2)^2 + 5

Step-by-step explanation:

The vertex form of a parabola is

y = a(x - h)^2 + k where (h, k) is the vertex.

Here h = -2 and k = 5 so the following could be the required equation\;

y = 3(x+2)^2 + 5

7 0

3 years ago

2x-3x²=5 in standard formation​

2x-3x²=5 in standard formation