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3 years ago

What is the pie root of 10,470

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

6 0

The pi root of 10,470 is approximately 321.457

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if angle 1 and angle 2 are vertical angles, angle 2 and angle 3 are complementary angles and the measure of angle 3 is 56 degree

dimulka [17.4K]

Answer:

34º

Step-by-step explanation: so basically you know a complementary angle is 2 angles that make a straight line and a vertical angle crosses which means 2 angles are congruent to each other so I drew a line that bisected the angle by drawing a line that cut through the right angle and i extended a line down from the right angle so now I have a vertical angle and a complementary so from here I labeled each angle <1 <2 <3 If I know angle 3 is 56º then I know 56 plus __ equals 90º I then subtracted 56 from 90 and I got 34 and since I am trying to figure out what angle 1 is I know that angle 1 and 2 are vertical angles therefore I know they are congruent to each other so <1 = 34º

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3 years ago

Which of the following is a polynomial? I NEED HELP ASAP*******

nadezda [96]

Answer:

7x⁷+2x⁻⁴+3 is a Polynomial

Step-by-step explanation:

I'm sorry I can't explain how I got the answer by hand. I graphed the Equation.

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3 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

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3 years ago

When I evaluate m=2 , h=1 then what would mh+3 would be?

Andrei [34K]

Answer:5

Step-by-step explanation: since you know that m=2, and h=1, mh is the same as m × h, or 2×1. Which is equal to 2, you then add 3 to get your answer

7 0

3 years ago

The result of adding a complex number to its conjugate is “an integer/a pure imaginary number/a real number/a whole number” and

sineoko [7]

Answer:

If Z is a complex number:

Z = a + b*i

where a and b are real numbers, and i is an imaginary number.

Then "a" is the real part.

"b*i" is the imaginary part.

The conjugate of Z is:

Zc = a - b*i

So the sign of the imaginary part changes.

Then:

Sum:

Z + Zc = (a + bi) + (a - bi) = 2*a + 0 = 2*a

and remember that a is a real number, then 2*a is also a real numer.

The correct answer is "A real number".

Difference:

Z - Zc = (a + bi) - (a - bi) = 2b*i

and this is a pure imaginary number, so here the correct answer is: "a pure imaginary number"

6 0

3 years ago