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Lapatulllka [165]
3 years ago
10

If Kate has an apple, an orange, a pear, a banana, and a kiwi at home and she wants to bring three pieces of fruit to school, ho

w many combinations of fruit can she bring?
a. 5!
b. 3!
c. 125
Mathematics
2 answers:
qaws [65]3 years ago
8 0

Step-by-step explanation:

i think A is correct

hope it helps

Zepler [3.9K]3 years ago
3 0
I think it’s A sorry if i’m wrong goodluck though
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What is true about the completely simplified sum of the polynomials 3x^2y^2-2xy^5 and -3x^2y^2 + 3x^2y^2 + 3x^4y?
WINSTONCH [101]

Given:

The polynomials are:

3x^2y^2-2xy^5

-3x^2y^2+3x^2y^2+3x^4y

To find:

The completely simplified sum of the polynomials.

Solution:

We have,

3x^2y^2-2xy^5

-3x^2y^2+3x^2y^2+3x^4y

The sum of given polynomials is:

Sum=3x^2y^2-2xy^5-3x^2y^2+3x^2y^2+3x^4y

Sum=-2xy^5+3x^4y+3x^2y^2

Therefore, the sum of the given polynomials is -2xy^5+3x^4y+3x^2y^2. It is a polynomial with degree 6 and leading coefficient -2.

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The sides of the angle are the lengths which it is formed
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If 4 cookies cost $4.00, how much does 1 cookie cost?
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Step-by-step explanation:

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(3+62) x (25-17) + 3*2 <br> 2 is an exponent btw
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Step-by-step explanation:

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3 years ago
For how many real values of x is <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B120-%5Csqrt%7Bx%7D%7D%20" id="TexFormula1" title
leva [86]

A good place to start is to set \sqrt{x} to y. That would mean we are looking for \sqrt{120-y} to be an integer. Clearly, y\leq 120, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since \sqrt{x} is a radical, it only outputs values from [0,\infty], which means y is on the closed interval: [0,120].

With that, we don't really have to consider y anymore, since we know the interval that \sqrt{x} is on.

Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval [0,120], which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:

\sqrt{120-\sqrt{x}}=k \implies \\ \sqrt{x}=k^2-120 \implies\\ x=(k^2-120)^2

Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.

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