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jasenka [17]
3 years ago
12

The question is below, please help

Mathematics
1 answer:
Alisiya [41]3 years ago
7 0
Im gonna go ahead and say it is 25



pls let me be right
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Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5
Marianna [84]
\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}

\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)

of course, you can always use  (x-1)(x-1)(2x+1)(x²-4x+5)  as well.
7 0
3 years ago
Divide 6 divided by 3/5 <br><br><br> Answer only if u know please:)
meriva

Answer:

The answer is 10 or 10/1

3 0
3 years ago
How do you simplify expressions with rational exponents
Rainbow [258]

Answer:

Step-by-step explanation:

Simplify expression with rational exponents can look like a huge thing when you first see them with those fractions sitting up there in the exponent but let's remember our properties for dealing with exponents. We can apply those with fractions as well.

Examples

(a)   (p^4)^{\dfrac{3}{2}}

From above, we have a power to a power, so, we can think of multiplying the exponents.

i.e.

(p^{^ {\dfrac{4}{1}}})^{\dfrac{3}{2}}

(p^{^ {\dfrac{12}{2}}})

Let's recall that when we are dealing with exponents that are fractions, we can simplify them just like normal fractions.

SO;

(p^{^ {\dfrac{12}{2}}})

= (p^{ 6})

Let's take a look at another example

\Bigg (27x^{^\Big{6}} \Bigg) ^{{\dfrac{5}{3}}}

Here, we apply the \dfrac{5}{3} to both 27 and x^6

= \Bigg (27^{{\dfrac{5}{3}}} \times x^\Big{\dfrac{6}{1}\times {{\dfrac{5}{3}}} }\Bigg)

= \Bigg (27^{{\dfrac{5}{3}}} \times x^\Big{\dfrac{2}{1}\times {{\dfrac{5}{1}}} }\Bigg)

Let us recall that in the rational exponent, the denominator is the root and the numerator is the exponent of such a particular number.

∴

= \Bigg (\sqrt[3]{27}^{5} \times x^{10} }\Bigg)

= \Bigg (3^{5} \times x^{10} }\Bigg)

= 249x^{10}

8 0
2 years ago
While shopping for a speaker, you see a discount.
harina [27]

Answer:

3*4. and 0.3*42.00 and 0.03*42

Step-by-step explanation:

4 0
2 years ago
6, 1, -4, -9,...
kykrilka [37]
Texaschic nailed it when saying that it's arithmetic

The recursive way to write this is to say

a_1 = 6
a_n = a_{n-1} - 5

which tells us "start at 6 and each time subtract off 5"

The a_n portion is the nth term while a_{n-1} is the term just before the nth term
3 0
3 years ago
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