2 years ago

Please help!!!! And show work please!

1 answer:

7 0

C. Table T
It is not a function if one x value gives more than 1 y-value. In table T, 3 gives 4 and -4 two different y values.

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How to solve (2.31*10^-6)+(5.87*10^-4)

S_A_V [24]

We want to calculate the following

$2.31\cdot10^{-6}+5.87\cdot10^{-4}$

Using the properties of exponents, we have that

$10^{-6}=10^{-4}\cdot10^{-2}$

So we have

$2.31\cdot10^{-6}+5.87\cdot10^{-4}=2.31\cdot10^{-2}\cdot10^{-4}+5.87\cdot10^{-4}$

So, if factor 10^-4 on the right side, we have

$10^{-4}(2.31\cdot10^{-2}+5.87)$

Note that

$2.31\cdot10^{-2}=0.0231$

Then,

$5.87+2.31\cdot10^{-2}=5.87+0.0231=5.8931^{}$

So we have that

$2.31\cdot10^{-6}+5.87\cdot10^{-4}=5.8931\cdot10^{-4}$

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Dennis_Churaev [7]

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Leno4ka [110]

Changing from negative to positive

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Line E'F' has endpoints located at E'(1,0) and F'(1,3). Line EF was dilated by a scale of 1/2 from the orgin. Which statement de

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Answer:

If line EF was dilated by a scale of 1/2 from the origin to get line E'F', then if line E'F' is dilated by a scale of 2 from the origin, EF is obtained. Also, the length of EF is double than the length of E'F'.

Dilation by a scale of 2 from the origin transforms point (x,y) into (2*x, 2*y)

E'(1,0) -> E(2,0)

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3 years ago

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