Simplify the following polynomial and write the answer in standard form. (-2x^3+5x^2-4x+8)-(-2x^3+2x-3)

1 answer:

4 0

$(-2x^3+5x^2-4x+8)-(-2x^3+2x-3)\\\\=-2x^3+5x^2-4x+8-(-2x^3)-2x-(-3)\\\\=-2x^3+5x^2-4x+8+2x^3-2x+3\\\\=(-2x^3+2x^3)+5x^2+(-4x-2x)+(8+3)\\\\=5x^2-6x+11$

You might be interested in

Find the value of x.

AfilCa [17]

Answer:

Step-by-step explanation:

5 0

3 years ago

An area of a square is 1.56 square feet what si the area of a floor covered with 10 to the third power tiles?

Diano4ka-milaya [45]

Answer: 1560 ft squared

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the greatest common factor for both terms: 24x + 8

Lyrx [107]

$24x=\boxed{2}\cdot\boxed{2}\cdot\boxed{2}\cdot3\cdot x\\\\8=\boxed{2}\cdot\boxed{2}\cdot\boxed{2}\\\\GCF(24x,\ 8)=\boxed{2}\cdot\boxed{2}\cdot\boxed{2}=8\\\\24x+8=8(3x+1)$

3 0

3 years ago

I have 2 triangle faces. i have 3 rectanglar faces. i have 6 vertices. i have 9 edges. what am i?

MissTica

Answer:

You are a rectangular prism.

Step-by-step explanation:

Rectangular prisms fill the criteria you described: two triangular faces, three rectangular faces, 6 vertices, and 9 edges. Thus, the answer is correct.

________________________________________________

Hoped that helped! If so, mark me Brainliest!

8 0

3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago