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3 years ago

(x^3 + y^3)/(x - y) divide the polynomials and show your work please.

Mathematics

1 answer:

UkoKoshka [18]3 years ago

6 0

Please see attached.

I hope this helps.

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Please help me with this...

AURORKA [14]

Answer:

- 84 + 10i

Step-by-step explanation:

A complex number in standard form is a ± bi

a is the real part and bi the imaginary part

and $\sqrt{-1}$ = i

Given

$\sqrt{-100}$ - 84

= $\sqrt{100(-1)}$ - 84

= $\sqrt{100}$ × $\sqrt{-1}$ - 84

= 10i - 84

= - 84 + 10i ← in standard form

3 0

3 years ago

I am extremely confused about the length of line segment AB. Could I get some help? (I have attached an image to this post)

Evgen [1.6K]

Answer:

The length of line segment AB is 9 units.

Step-by-step explanation:

To find this, you count how many squares there are between the two points. Hope I helped.

7 0

3 years ago

Convert the unit. of length

Sergio [31]

Answer:

Step-by-step explanation:

<em>thats all i got sorry if thats wrong but i hope this helps.</em>

4 0

2 years ago

Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0

Debora [2.8K]

So, the definite integral $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Given that

$\int\limits^1_0 {x^{2} } \, dx = 13$

We find

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx$

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, $Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6\int\limits^1_0 {x^{2} } \, dx$