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Olin [163]
3 years ago
15

9

x} ) ^5 " align="absmiddle" class="latex-formula"> How do you rewrite the following radical expression in rational exponent form?
Mathematics
2 answers:
BlackZzzverrR [31]3 years ago
8 0
<span>How do you rewrite the following radical expression in rational exponent form?
</span>9 1/2 is the answer
Maksim231197 [3]3 years ago
3 0
9 1/2 lol you're welcome I can see you and I hate him
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5,250 as scientific notation
earnstyle [38]

\text{Hey there!}

\text{5.250 as a scientific notation}

\bf{The\ answer\ is:}\boxed{\bf{5.25 \times10^3}}}}

\text{Because, we had to move your decimal approximately 3 placecs to the left}

\rightarrow\tex{= 5250. }

\boxed{\text{That's how we know the answer is: \bf{\boxed{5.25 \times10^3}}}}\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

4 0
3 years ago
Read 2 more answers
For f(x) = 3x + 1 and g(x) = x2 - 6, find (f+ g)(x).
kirza4 [7]

Answer:

B

Step-by-step explanation:

(f+g)(x)=f(x)+g(x)

3x+1+x2-6 = x2+3x-5

5 0
3 years ago
Prove that an = 4^n + 2(-1)^nis the solution to
olga nikolaevna [1]

Answer:

See proof below

Step-by-step explanation:

We have to verify that if we substitute a_n=4^n+2(-1)^n in the equation a_n=3a_{n-1}+4a_{n-2} the equality is true.

Let's substitute first in the right hand side:

3a_{n-1}+4a_{n-2}=3(4^{n-1}+2(-1)^{n-1})+4(4^{n-2}+2(-1)^{n-2})

Now we use the distributive laws. Also, note that (-1)^{n-1}=\frac{1}{-1}(-1)^n=(-1)(-1)^{n} (this also works when the power is n-2).

=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}

=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}

=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n

then the sequence solves the recurrence relation.

4 0
3 years ago
Sin40°-sin20°/cos220°-cos200=√3​
Mumz [18]

Answer:

1.7320508

Step-by-step explanation:

7 0
3 years ago
Multiply (x-2) (2x+4)
Bas_tet [7]

Answer:

2x {}^{2}  - 8 = 0

Step-by-step explanation:

(x-2) (2x+4)  = 0\\ 2x {}^{2}   +  4x - 4x - 8 = 0 \\ 2x {}^{2}  - 8 = 0

5 0
3 years ago
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