All categories

2 years ago

Jolly measured the cake. Then she calculated that it has a circumference of 15cm. what is the cake's diameter? Use 3.14 for pi.

Mathematics

1 answer:

Archy [21]2 years ago

8 0

Answer:

diameter = circumference / pi

circumference = 15cm

diameter = 15 / 3.14

= 4.77707006369 cm

hope this answer helps you.....

please mark as brainliest... thank you!

You might be interested in

4 1/4 divided by 2/7 (x+8)-(3x- 5) Explain how you got the answer!!!!!

Dmitry_Shevchenko [17]

Turn it to improper fractions.

4 1/4 divided by 2/7

Divide them.

17/4 divided by 2/7

17 2 17 x 7

--- ÷ --- = --------- = 119/8 OR 14 7/8

4 7 4 x 2

-----------------------------------------------------

Distribute to take out the parentheses.

(x + 8) - (-3x - 5)

Combine like terms.

x + 8 + 3x + 5

Answer.

4x + 13

8 0

2 years ago

Find the composite area of the shaded region

Anvisha [2.4K]

Answer:

what region

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

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

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