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notka56 [123]
3 years ago
10

Can you help me simplify these double radicals?

Mathematics
1 answer:
Montano1993 [528]3 years ago
8 0
You know that
.. (a +b)^2 = a^2 +b^2 +2ab
If these are square roots, then you have
.. (√a +√b)^2 = a +b +2√(ab)
.. (√a -√b)^2 = a +b -2√(ab)

In each case, you can put the inner radical in the form 2√something, then look for two factors of "something" that add to the other constant.
The result is then √factor1 ±√factor2, with the sign matching that under the original radical.

1. 1 +√5
2. 2 +√2
3. √3 +√7
4. √5 -√2 . . . . . . . . Technically, √(x^2) = |x|. By expressing the result as a positive number (smaller subtracted from larger), we don't have to worry about absolute value.

5. See 3 and 4. √7 -√3
6. √20 = 2√5. See 1 and 4.
.. √5 -1
7. 2 -√3
8. 4√3 = 2√12 = 2√(6*2)
.. √6 +√2
9. 6√3 = 2√(9*3)
.. 3 -√3
10. 8√3 = 2√(16*3)
.. 4 +√3
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xenn [34]

Answer:

(x,y) => (x+3, y+4)

Step-by-step explanation:

Given a triangle DEF, D(4,2), E(3,3), F(2,1).

Centroid of the area (and the vertices) equals the mean of the coordinates, namely ( (4+3+2)/3, (2+3+1)/3 ) = (3,2)

To translate (3,2) to (6,6), we need the rule

(x,y) => x+(6-3), y+(6-2), or

(x,y) => (x+3, y+4)

5 0
3 years ago
If f(x) = <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D" id="TexFormula1" title="\frac{1}{9}" alt="\frac{1}{9}" align=
boyakko [2]

Answer:

Solution given:

f(x) = \frac{1}{9} x-2

let f(x)=y

y = \frac{1}{9} x-2

interchanging role of x and y

x= \frac{1}{9} y-2

x+2= \frac{1}{9} y

y=9(x+2)

y=9x+18

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<u>a</u><u>n</u><u>d</u>

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6 0
3 years ago
Simplify the problem given.
maks197457 [2]
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Done! :)
4 0
3 years ago
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julsineya [31]
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Begin by combining like terms:
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The equation is now simplified.
Now, it is time to find the value of the variable.

-2e - 5 = 15

Step 1: Add 5 to both sides.

We now have -2e = 20.

Step 2: Divide both sides by -2 to isolate the variable.

-2e / 2 = e; 20 / -2 = -10.

Therefore, e = -10.
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ankoles [38]

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Answer:

Option A

5 0
3 years ago
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