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Liula [17]
3 years ago
11

What is the image point of (-4, 8) after a translation right 4 units and up 2 units?

Mathematics
1 answer:
Vitek1552 [10]3 years ago
4 0

Given:

Point = ( -4,8).

To find:

The image point of (-4, 8) after a translation right 4 units and up 2 units.

Solution:

We know that, if a figure translated 4 units right and 2 units up, then the rule of translation is

(x,y)\to (x+4,y+2)

The point is (-4,8).

Putting x=-4 and y=8, we get

(-4,2)\to (-4+4,8+2)

(-4,2)\to (0,10)

Therefore, the image of point (-4,8) is (0,10).

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Step-by-step explanation:

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3 years ago
A gas occupies 1.00 L at standard temperature. What is the volume at 606 K?
tino4ka555 [31]

Answer:

the volume at 606k is 2.22....the third one

Step-by-step explanation:

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3 years ago
Simplify radicals problems attached please help!
Finger [1]

Answer:

  G7.  (2√3)/3

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Step-by-step explanation:

G7.

\dfrac{2}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}

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G8.

\dfrac{3}{2 +\sqrt{7}}=\dfrac{3}{2+\sqrt{7}}\cdot\dfrac{2-\sqrt{7}}{2-\sqrt{7}}=\dfrac{6-3\sqrt{7}}{2^2-(\sqrt{7})^2}\\\\=\dfrac{6-3\sqrt{7}}{-3}=\sqrt{7}-2

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G9.

\dfrac{2-\sqrt{3}}{3-\sqrt{2}}=\dfrac{2-\sqrt{3}}{3-\sqrt{2}}\cdot\dfrac{3+\sqrt{2}}{3+\sqrt{2}}=\dfrac{(2-\sqrt{3})(3+\sqrt{2})}{9-2}\\\\=\dfrac{6+2\sqrt{2}-3\sqrt{3}-\sqrt{6}}{7}

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<em>Comment on the problems</em>

In most cases, these expressions are the simplest possible (take the least amount of ink to draw, and take the fewest math operations to evaluate). What seems to be intended is that the denominator be made a rational number. This is done by multiplying the given fraction by a fraction equal to 1 that has the same denominator but with the sign of the radical reversed (unless, as in the first case, the radical is by itself).

The purpose of doing this is to take advantage of the fact that (a-b)(a+b) = a²-b², so if "a" or "b" is a square root, that root will not be seen in the product. In problem G9, we see this can make the numerator quite messy--not exactly a simpler form--but all the irrational numbers are in the numerator.

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