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2 years ago

13

A park has a large lot there is grass on 2/3 of the park the picnic area covers 1/5 of the grass area what fraction of the park

is covered by the picnic

1 answer:

uranmaximum [27]2 years ago

3 0

Answer:

3 2/3

Step-by-step explanation:

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How does Reflection work in real-life situations?

bazaltina [42]

Answer:

reflection is defined as the change in the direction of a wavefront at the interface between two different media, bouncing the wavefront back into the original medium. A common example of reflection is reflected light from a mirror or a still pool of water, but reflection affects other types of waves beside light.

Step-by-step explanation:

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2 years ago

Read 2 more answers

What is 100 x 0.95=_____

kifflom [539]

Answer: 95

Step-by-step explanation:

That’s the same as 0.95 of 100. That’s because 0.95 represents 0.95 and 100 represents 100, the of is the same as x. You can switch the two numbers and you can still get the same answer. So, 0.95 of 100 is 95.

5 0

2 years ago

What is the output of this function when the input is 2

nalin [4]

Easy question.

So, the slope is -x and the y-intercept is at -4.

y = -x - 4

Use this equation and you will find:

y = -2 - 4

y = -6

Answer: (2, -6)

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2 years ago

Determine the value of x in the diagram where lines a and b are parallel

Yakvenalex [24]

If lines a and b are parallel, the two angles given would be the same.
2x - 5 = x + 20
x = 25

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