-f+2+4f=8-3f solve for f

Mathematics

2 answers:

Mashutka [201]3 years ago

6 0

Answer:

Step-by-step explanation:

Brrunno [24]3 years ago

4 0

Answer:

Step-by-step explanation:

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A scale on a map shows that 2 inches equals 25 miles how many inches on the map represents 60 miles?? I NEED HELP ASAP

irakobra [83]

Answer:

60 miles is equal to 4.8 inches.

Step-by-step explanation:

To find the answer to this question, we need divide 60 by 25 to find out how much to multiply 2 by so its equal.

60/25=2.4

2*2.4=4.8

60 miles is equal to 4.8 inches.

Hope this helps!

6 0

3 years ago

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Find the circumference and area of a circle with a diameter of 14 inches. Leave your answers in terms of pi. C = 14π; A = 49π. C

Olenka [21]

Given:
diameter = 14 inches
radius = 14 inches / 2 = 7 inches

Circumference of a circle = 2 π r = 2 * π * 7inches = 14 inches * π

Area of a circle = π r² = π * (7in)² = π * 49in²

C = 14π and A = 49π

5 0

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

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