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2 years ago

Apply the square root principle to solve (x – 2)2 + 20 = 0.

Mathematics

1 answer:

Temka [501]2 years ago

7 0

(x-2)^2+20=0

(x-2)^2=-20

If you root both sides you will yield complex numbers (i, imaginary unit is such that i^2=-1),

x-2=20i

x=2+20i

and whenever x is a solution a complex conjugate x* is also a solution,

x*=2-20i

So the answer is A.

Hope this helps :)

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If a central angle measures 25, then the measure of its intercepted arc is

borishaifa [10]

Medida do angulo central = medida do arco interceptado
arco = 25º e ângulo central = 25º

7 0

3 years ago

What is 5% of $30?<br>$150<br>$35<br>$1.50<br>$25

fomenos

Answer:

$1.50

Step-by-step explanation:

5% of 30 is 1.5 AKA 1.50

All you have to do is remove then add the dollar sign again!

Hope this helps! :)

6 0

3 years ago

Read 2 more answers

What’s the correct answer

n200080 [17]

The third one would be the correct answer. I separately multiplied each number in the first section by each number in the second section. For instance, 0 x 2 is 0, 4 x 5 is 20, 5 x 3 is 15, and 2 x 0 is 0. All of these numbers are in the third answer.

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2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

2 years ago

This is a bit confusing for me, can anyone help me out?

AnnZ [28]

Answer:

circumference=471

area=17671

Step-by-step explanation:

circumference=150pi=150*3.14=471

area=150/2=75 and 75^2pi=5625pi=17671

4 0

2 years ago