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2 years ago

7

A square is inscribed in a circle. A tangent is drawn to the vertex. What size angle is formed? Select one: a. 45 b. 180 c. 22.5

d. 90

1 answer:

olga nikolaevna [1]2 years ago

8 0

Answer:

I believe its A

Step-by-step explanation:

You might be interested in

Write in standard form

Natalka [10]

Answer:

2.988*10^8

Step-by-step explanation:

to do standard form you have to move the decimal to after the first number so in this case we would move the decimal 8 places from 298800000.0

to 2.98800000 and we can remove the zeros to make it 2.988

then you will plug the times ten and because we move the decimal over 8 it would be 10 to the power of 8 so your answer would be 2.988*10^8

hope this helps have a good day

<em>brainlest please? :)</em>

5 0

2 years ago

Read 2 more answers

Answer it plzzz plz <br>you would live long if correct answer

Ede4ka [16]

Answer: 3^15

Step-by-step explanation:

Explained in attached picture. Hope you'll find it easy to understand. If not, you can comment :)

8 0

2 years ago

Which set of shapes could be used to form a net for a square pyramid?

bulgar [2K]

Answer:

A.1 square and 4 triangles

Step-by-step explanation:

A pyramid has sides that are triangular faces and a base. In a square pyramid, the base is a square.

The net therefore has 1 Square and 4 Triangles.

A net is given in the attached diagram:

8 0

3 years ago

What is the area of this compound shape?

Paraphin [41]

Answer:

139.36

Step-by-step explanation:

first you do 8*8 because that would be the area of the square part

8*8=64

then find the area of the semicircle

first the Radis of it is 4 because 8/2=4

A=3.14R^2/2

3.14*4^2/2

3.14*16/2

50.24/2

25.12

now because there are 3 of the semicircles you do

25.12*3

75.36

then you add the 64 from the square to get your answer

75.36+64

139.36

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