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3 years ago

A triangle has two sides that measure 4.56 cm and 8.65 cm. Which could be the measure of the third side?

Mathematics

1 answer:

Julli [10]3 years ago

4 0

4.56 because it said that the triangle said it had two sides that measure 4.56

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The perimeter of a college athletic field is 96 meters and the length is 16 m more than the width. Find the length and width.

Andre45 [30]

In order to figure this out, you are going to do the math backwards. If the perimeter of the entire field is 96 meters, you will want to remove 16, which will give you 80 meters. You will then divide this by 2, giving you 40 meters. The width of the field will be 40 meters. That means that the length of the field is going to be 56 meters.

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2 years ago

I have 4 millions, 8 thousands, 11 tens, 6 ones, 9 tenths and 5 hundredths. what number am I ?

egoroff_w [7]

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2 years ago

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The sum of two numbers is 66. Two times the smaller number is 15 more than the larger number. What are the numbers?

Rudiy27

Here is the set up:

Let x and y be the two numbers with x the smaller and y the larger number.

x + y = 66

2x = y + 15

We have two equations in two unknowns.

Take it from here.

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2 years ago

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write a possible point slope form of the equation of the line that passes through the points (4,4) and (3,-2)?

Mashutka [201]

m=-2-4/3-4=-6/-1=6

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y=6x-20

6 0

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