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Alenkasestr [34]
3 years ago
12

Plz help also show your work.✌

Mathematics
1 answer:
I am Lyosha [343]3 years ago
5 0
The answer is A because c is the amount of cookies and she ate three. You divide it by two because she only gave HALF of it away.
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Choose one <br> Help me please
AfilCa [17]

Answer:

A

Step-by-step explanation:

Smallest to Largest

Natural numbers

Whole numbers

Integers

Rational numbers

4 0
3 years ago
Suppose you like to keep a jar of change on your desk. Currently, the jar contains the following: 22 Pennies 27 Dimes 9 Nickels
kap26 [50]

Answer:

P(Penny\ and\ Dime) = \frac{9}{116}

Step-by-step explanation:

Given

Pennies = 22

Dimes = 27

Nickels = 9

Quarters = 30

Required

P(Penny\ and\ Dime)

This is calculated as:

P(Penny\ and\ Dime) = P(Penny) * P(Dime)

Since it is a selection without replacement, the computation is:

P(Penny\ and\ Dime) = \frac{Penny}{Total} * \frac{Dime}{Total-1}

So, we have:

P(Penny\ and\ Dime) = \frac{22}{22+27+9+30} * \frac{Dime}{22+27+9+30-1}

P(Penny\ and\ Dime) = \frac{22}{88} * \frac{27}{87}

P(Penny\ and\ Dime) = \frac{1}{4} * \frac{9}{29}

P(Penny\ and\ Dime) = \frac{9}{116}

5 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0
2 years ago
Which of the following is a quadratic function?
sp2606 [1]

A is the answer, because the highest exponent is 2.

<u>quadratic</u>

A's expression foiled would be 2x^2 + x - 6

B's expression foiled would be 3x - 12, where the exponent on x is only 1.

D's expression foiled would be x^3 - x^2 - 12x, where the exponent on x is 3.

6 0
3 years ago
What is 2×(-5)-(-8)=? cause I need the answer for something rlly bad
shepuryov [24]
The answer to this is -2
6 0
3 years ago
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