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3 years ago

How many terms does the polynomial have?

Mathematics

2 answers:

8090 [49]3 years ago

5 0

<h2>Hello!</h2>

The answer is:

The polynomial have 3 terms. (it's a trinomial).

A polynomial is an expression which consists of one or more terms (numbers or variables) being added or subtracted.

So, we are given the polynomial:

$x^{2} +xy-y^{2}$

We have that there are three terms separated by differents being added and subtracted, so, the polynomial has 3 terms, and it's a trinomial.

Have a nice day!

AveGali [126]3 years ago

3 0

3 terms. Just count the terms (which are separated by the +/- signs)

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Answer:

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3 years ago

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The total surface area of this cuboid is 112cm^2. Find the value of x.

Oxana [17]

Answer:

The value of x is 3

Step-by-step explanation:

Let us study the face of the cuboind

∵ The cuboid has 6 rectangular faces

∵ Each opposite faces area equal in areas

∴ 2 faces of dimensions 10 cm and 2 cm

∴ 2 faces of dimensions 10 cm and x cm

∴ 2 faces of dimensions 2 cm and x cm

∵ The total surface area of the cuboid is the sum of the areas of the 6 faces

∵ The area of the rectangle = length × width

∴ The total surface area = 2(10 × 2) + 2(10 × x) + 2(2 × x)

∴ The total surface area = 2(20) + 2(10x) + 2(2x)

∴ The total surface area = 40 + 20x + 4x

→ Add the like terms 20x and 4x

∴ The total surface area = 40 + 24x

∵ The total surface area of this cuboid is 112 cm²

→ Equate the two sides of the total surface area

∴ 40 + 24x = 112

→ Subtract 40 from both sides

∵ 40 - 40 + 24x = 112 - 40

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4 years ago

find a homogeneous linear system of two equations in three unknowns whose solution space consists of those vectors in R3 that ar

Leona [35]

$\mathbf a=(1,1,1)^\top=1\,\mathbf i+1\,\mathbf j+1\,\mathbf k$
$\mathbf b=(-2,3,0)^\top=-2\,\mathbf i+3\,\mathbf j$
$\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&1&1\\-2&3&0\end{vmatrix}=-3\,\mathbf i-2\,\mathbf j+5\,\mathbf k=(-3,-2,5)^\top$

Basically, you're looking for a matrix $\mathbf A$ such that

$\mathbf A(\mathbf a\times\mathbf b)=\mathbf 0$

i.e. a matrix $\mathbf A$ whose nullspace with basis vector $\mathbf a\times\mathbf b$ .

By the rank-nullity theorem, the rank of $\mathbf A$ and the dimension of its nullspace must add up to the number of columns, so

$\mathrm{rank}\mathbf A+\underbrace{\mathrm{null}\mathbf A}_1=3\implies\mathrm{rank}\mathbf A=2$

One easy choice for a row would be $\begin{bmatrix}1&1&1\end{bmatrix}$ , since

$(1,1,1)(-3,-2,5)^\top=0$

Now you only need to find another combination such that the second row of $\mathbf A$ is independent of the first. An easy choice for this is to let the first element be 0, and the next be 1. Then the last element must be $\dfrac25$ , as

$\left(0,1,\dfrac25\right)(-3,-2,5)^\top=0$

So,

$\underbrace{\begin{bmatrix}1&1&1\\0&1&\frac25\end{bmatrix}}_{\mathbf A}\begin{bmatrix}-3\\-2\\5\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

is one possible solution.

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3 years ago