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2 years ago

Real numbers can be written in the form a b, where a and b are integers and b≠0, are called __________.

Mathematics

1 answer:

pashok25 [27]2 years ago

3 0

Answer:

Rational number

Step-by-step explanation:

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Find the midpoint of the line segment joining the points P1 and P2.

ludmilkaskok [199]

Formula for midpoint between two points is M(x,y)

x=(x1+x2)/2 and y=(y1+y2)/2

In our case (x1,y1)=(m,b) and (x2,y2)=(0,0)

x=(m+0)/2=m/2 and y=(b+0)/2=b/2 M(m/2,b/2)

Good luck!!!

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3 years ago

Find the diameter of each circle.

Vedmedyk [2.9K]

Answer:

Diameter Length: ( About ) 5.4 km; Option B

Step-by-step explanation:

~ Let us apply the Area of the Circle formula πr^2, where r ⇒ radius of the circle ~

1. We are given that the area of the circle is 22.9 km^2, so let us substitute that value into the area of the circle formula, solving for r ( radius ) ⇒ 22.9 = π * r^2 ⇒ r^2 = 22.9/π ⇒ r^2 = 7.28929639361.... ⇒

radius = ( About ) 2.7

2. The diameter would thus be 2 times that of the radius by definition, and thus is: 2.7 * 2 ⇒ ( About ) 5.4 km

Diameter Length: ( About ) 5.4 km

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3 years ago

15 times a number increased

Minchanka [31]

Answer:

Step-by-step explanation:

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2 years ago

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For a school fund-raiser, 10 students sold a total of 90 boxes of cookies. Which of the following can be calculated from this in

allsm [11]

Answer: what of the following

Step-by-step explanation:

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3 years ago

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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

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

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2 years ago