2 years ago

6. Chapter 6 Quiz 2 (Lessons 6-3 and 6-4)

Mathematics

1 answer:

Firlakuza [10]2 years ago

5 0

Answer:

What in the world is this even for?

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This is a practice test but I don’t understand stand this one pls help fast

Kisachek [45]

Step-by-step explanation:

To transform a function vertically, add 3 to the constant.

$5 + 3 = 8$

So the function is

$3 {x}^{2} + 8$

Graph the function in desmos to show the translation vertically if you want proof

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1 year ago

suppose a radioactive substance decays according to the equation y=200e^-0.75t, where y represent the mass in gram after t hours

Daniel [21]

Answer:

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Step-by-step explanation:

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2 years ago

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belka [17]

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2 years ago

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Answer?...............

____ [38]

Answer:

-9 1/4

Step-by-step explanation:

-5 3/4 = -23/4

3 1/2 = 7/2

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3 years ago

Which expression is equivalent to RootIndex 3 StartRoot 32 x Superscript 8 Baseline y Superscript 10 Baseline EndRoot?

evablogger [386]

Answer:

The equivalent expression to the given expression is $\sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}$

Step-by-step explanation:

The given expression is $\sqrt[3]{32x^8y^{10}}$

To find the equivalent expression:

$\sqrt[3]{32x^8y^{10}}=(32x^8y^{10})^{\frac{1}{3}}$

We may write the above expression as below:

$=(32^{\frac{1}{3}})((x^8)^{\frac{1}{3}})((y^{10})^{\frac{1}{3}})$

$=(2)((4)^{\frac{1}{3}})(x^6)\times (x^2)(x^{\frac{2}{3}})(y^3)(y^{\frac{1}{3}})$ (using square root properties)

$=(2\sqrt[3]{4})(x^2\sqrt[3]{x^2})(y^3\sqrt[3]{y})$ (combining the like terms and doing multiplication )

$=2x^2y^3\sqrt[3]{4x^2y}$

Therefore $\sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}$

Therefore the equivalent expression to the given expression is $\sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}$

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3 years ago

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