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3 years ago

When adding polynomials, you combine like terms. Then put the polynomial in standard form. true or false

Mathematics

1 answer:

UkoKoshka [18]3 years ago

8 0

Answer:

True

Step-by-step explanation:

Polynomials are expressions containing one or more variables that are raised to powers beyond 2.

In standard form, a polynomial may be arranged based on the power of the variable. For example

X^3 + 2x^2 + 5

The above polynomial has been arranged in standard form. To add or subtract two polynomials, the two expressions are rearranged to combined like terms and then organized in standard form

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Is g(x)=-2f(x)+4 a vertical shift?

AnnyKZ [126]

Yes. It is a vertical shift up by 4 units.

A vertical shift will be up or down along the y-axis.

8 0

4 years ago

PLS someone help me im trying to get a 80% so i can finish this!

klasskru [66]

Answer:

any number less than 3

Step-by-step explanation:

3 0

3 years ago

The length of the top of a table is 3 m greater than the width. The area is 70 m^2. Find the dimensions of the table.

Keith_Richards [23]

Let's call the width $x$ . Since the length is three more than the width, it is $x+3$ .

The area (which we know to be 70) is given by the multiplication of width and length:

$x(x+3) = 70 \iff x^2+3x = 70 \iff x^2+3x-70 = 0$

To solve this equation, you can use the usual quadratic formula: given an equation $ax^2+bx+c=0$ , the two solutions are

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

which in your case becomes

$x_{1,2} = \frac{-3\pm\sqrt{9+280}}{2} = \frac{-3\pm 17}{2}$

So, two solutions are

$\frac{-3-17}{20} = -10,\qquad \frac{-3+17}{2} = 7$

Since we can't accept a negative length, we only accept the second solution.

So, the dimensions are 7 and 10

5 0

4 years ago

Read 2 more answers

Help please!!! Hurry

rosijanka [135]

Answer:

No solution

Step-by-step explanation:

4 0

3 years ago

In Which Quadrant is this true

34kurt

Given:

$\sin \theta$

$\tan \theta$

To find:

The quadrant in which $\theta$ lie.

Solution:

Quadrant concept:

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II, only $\sin\theta$ and $\csc\theta$ are positive.

In Quadrant III, only $\tan\theta$ and $\cot\theta$ are positive.

In Quadrant IV, only $\cos\theta$ and $\sec\theta$ are positive.

We have,

$\sin \theta$

$\tan \theta$

Here, $\sin\theta$ is negative and $\tan\theta$ is also negative. It is possible, if $\theta$ lies in the Quadrant IV.

Therefore, the correct option is D.

5 0

3 years ago