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LuckyWell [14K]
3 years ago
14

872,000,000,000 in standard form

Mathematics
2 answers:
rewona [7]3 years ago
8 0
The answer would be 8.72^11
BabaBlast [244]3 years ago
5 0

Answer:

eight hundred and seventy-two trillion

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Multiply and simplify: (6x + 3y)(6x − 3y)
nasty-shy [4]

Answer:

36x² - 9y²

Step-by-step explanation:

We see that the given expression (6x + 3y)(6x - 3y) looks just like the expression (a + b)(a - b). When multiplied out, this is a difference of squares identity that I highly recommend you memorise:

(a + b)(a - b) = a² - b²

Here, a = 6x and b = 3y, so plug these in:

(6x + 3y)(6x - 3y) = (6x)² - (3y)² = 36x² - 9y²

5 0
3 years ago
Read 2 more answers
Draw a net for each of the following shapes:
Margarita [4]
A. You would start off by a rectangle in the middle, then you would place to squares on the bottom and top if the rectangle. After you've done that, you place two rectangles (Same size as the original) on both sides of the original rectangle. Finally you add a rectangle (Same size) to either ends of the secondary rectangles. It should look like this.


B. You would start off with a rectangle in the middle, then you would add 2 equilateral triangles to the top and bottom of the rectangle. After that, you Simply put 2 rectangles (Same size as the original) on both sides of the rectangle. It should look something like this.

C. To make a pyramid, it's actually quite simple. You would start off with a square in the middle, and then place equilateral triangles on ALL sides of the square. It should look something like this.

I hope this helped ^^

4 0
3 years ago
ravis gas tank is 1/2 full .after he buys 2 gallons of gas it is 2/3 full how many gallons can ravis tank hold?
aliya0001 [1]
So
1/2x+2=2/3x
convert to common fraction
1/2=3/6
2/3=4/6
3/6x+2=4/6x
subtract 3/6x from both sides
2=1/6x
multiply boh tisdes by 6
12=x
the tank can hold 12 gallon

8 0
3 years ago
Which of these limits evaluate to 0?
Vikentia [17]
<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\

----------

Part II

\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\

----------

Part III

\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\

7 0
2 years ago
Hey could you please help me and my new friend found out this answer?
Ludmilka [50]

Answer:

i will be you friend and help you but where the question

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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