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Romashka-Z-Leto [24]
4 years ago
13

How many zeros does the function f(x) = 4x11 − 20x7 + 2x3 − 15x + 14 have?

Mathematics
2 answers:
stepan [7]4 years ago
8 0
There are 3 zeros in this function
satela [25.4K]4 years ago
7 0

recall the fundamental theorem of algebra, there are as many zeros as the degree of the polynomial.


\bf f(x)=\stackrel{\textit{the polynomial is of the 11th degree}}{4x^{\stackrel{\downarrow }{11}}-20x^7+2x^3-15x+14}

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3 years ago
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$648 increased 2/3 itself equals what amount?
ladessa [460]

Answer:

1080

Step-by-step explanation:

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3x^2-4x+1, match the following
il63 [147K]

Answer:

x = variable

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8 0
3 years ago
Mr. Snow bought 90 grams of Christmas candy for each of his 14 grandchildren. How many total kilograms of candy did he buy?
nignag [31]

Answer:

1.26 kg.

Step-by-step explanation:

We have been given that Mr. Snow bought 90 grams of Christmas candy for each of his 14 grandchildren. We are asked to find the amount of candy bought by Mr. Snow in kilograms.

First of all, we will find the amount of candy in grams by multiplying 90 grams by 14 as:

\text{Amount of candy bought in grams}=14\times 90

\text{Amount of candy bought in grams}=1260

We know that 1 kilogram is equal to 1000 grams. To convert 1260 grams into kg, we will divide 1260 by 1000 as:

\text{Amount of candy bought in kilograms}=\frac{1260}{1000}

\text{Amount of candy bought in kilograms}=1.26

Therefore, Mr. Snow bought 1.26 kilograms of candy.

7 0
3 years ago
Can someone please answer this for me i cant figure it out.
Jlenok [28]
<h3>Answer:</h3>

\displaystyle x^{\frac{2}{3}}

<h3>Step-by-step explanation:</h3>

The rules of exponents tell you ...

... (a^b)(a^c) = a^(b+c) . . . . . . applies inside parentheses

... (a^b)^c = a^(b·c) . . . . . . . . applies to the overall expression

The Order of Operations tells you to evaluate inside parentheses first. Doing that, you have ...

... x^(4/3)·x^(2/3) = x^((4+2)/3) = x^2

Now, you have ...

... (x^2)^(1/3)

and the rule of exponents tells you to multiply the exponents.

... = x^(2·1/3) = x^(2/3)

3 0
3 years ago
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