Which of these can help you become a better test-taker?

1 answer:

5 0

Personally I would say C because when you take good notes you can go over what you don’t remember. You can also use your notes as a study guide.

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The length of human pregnancies from conception to birth varies according to a distribution that is approximately normal with a

Oduvanchick [21]

218, 314 are the two values do the middle 95% of the lengths of all pregnancies fall.

8 0

2 years ago

Explain Pe^(rt) How do I use this formula ?

Montano1993 [528]

Answer:

The equation for "continual" growth (or decay) is A = Pert, where "A", is the ending amount, "P" is the beginning amount (principal, in the case of money), "r" is the growth or decay rate (expressed as a decimal), and "t" is the time (in whatever unit was used on the growth/decay rate).

Step-by-step explanation:

<h2> Don't sweat here is a video link too </h2>

Compounding Continuously Pert Formula

https://youtu.be/dFsBfi9W7sQ

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2 years ago

(ASAP PICTURE ADDED) What is the simplified form of the following expression?

bonufazy [111]

Answer:

option c is correct.

Step-by-step explanation:

$7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)$

WE need to simplify this equation.

Solve the parenthesis of each term.

$=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right$

Now, We will find factors of the terms inside the square root

factors of 2: 2

factors of 16 : 2x2x2x2

factors of 8: 2x2x2

Putting these values in our equation: $=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)$

Adding like terms we get:

$=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\$