Please help i dont get this at all so please i will highly appreciate it .Thank youuu!!!

Factorize completely the expression(h+2k)²+4k²-h²

julsineya [31]

The expression factorized completely is (h +2k)[(h+2k) + (2k-h)]

From the question,

We are to factorize the expression (h+2k)²+4k²-h² completely

The expression can be factorized as shown below

(h+2k)²+4k²-h² becomes

(h+2k)² + 2²k²-h²

(h+2k)² + (2k)²-h²

Using difference of two squares

The expression (2k)²-h² = (2k+h)(2k-h)

Then,

(h+2k)² + (2k)²-h² becomes

(h+2k)² + (2k +h)(2k-h)

This can be written as

(h+2k)² + (h +2k)(2k-h)

Now,

Factorizing, we get

(h +2k)[(h+2k) + (2k-h)]

Hence, the expression factorized completely is (h +2k)[(h+2k) + (2k-h)]

Learn more here:brainly.com/question/12486387

5 0

2 years ago

Read 2 more answers

Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

4 0

2 years ago

This is also a rational and irrational worksheet so which one would it be?

Elden [556K]

If you calculated it you would get 20.25 so it is irrational

5 0

3 years ago

I need heelp<br> (-6)-(+8)=

zhuklara [117]

Answer:

I believe its 2 so sorry if im wrong

Step-by-step explanation:

4 0

3 years ago

Find the simple interest for $3200 invested at 7% for 2 years.

Artemon [7]

Answer:

$448

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 7%/100 = 0.07 per year,

then, solving our equation

I = 3200 × 0.07 × 2 = 448

I = $ 448.00

USE THE EQUATION Interest= Principal x rate x time

3 0

2 years ago