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4vir4ik [10]
3 years ago
9

Find the perimeter of the polygon. Assume that lines which appear to be tangent are tangent. Round to the nearest tenth if neces

sary.

Mathematics
1 answer:
ikadub [295]3 years ago
4 0

The perimeter of the polygon is 53.6 units.

Solution:

The reference image to the answer is attached below.

AP = 8, BQ = 7.8 and CR = 11

AP and AR are tangents to a circle from an external point A.

BP and BA are tangents to a circle from an external point B.

CQ and CR are tangents to a circle from an external point C.

<em>Tangents drawn from an external point to a circle are equal in length.</em>

⇒ AP = AR, BP = BQ and CQ = CR

AR = 8

BP = BQ

⇒ BP = 7.8

CQ = CR

⇒ CQ = 11

Perimeter of the polygon = AP + BP + BQ + CQ + CR + AR

                                          = 8 + 7.8 + 7.8 + 11 + 11 + 8

                                          = 53.6

The perimeter of the polygon is 53.6 units.

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Solve for x I need help on this question
AlexFokin [52]

Answer:

x\approx7.8

Step-by-step explanation:

In relation to the given angle, we are given the triangle's opposite side and hypotenuse. Therefore, we use the sine function to set up a proportion and solve for the opposite side:

sin(\theta)=\frac{opposite}{hypotenuse}

sin(23^\circ)=\frac{x}{20}

20sin(23^\circ)=x

x=20sin(23^\circ)

x\approx7.8

Therefore, the length of the opposite side is about 7.8 units

3 0
3 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
3 years ago
On farm there are 200 cows from those 200 cows there are 150 cows that can give milk .in one corral there are 20 cows. How many
LiRa [457]

Here the two samples should be comparable .

The total number of cows in the farm = 200

the total number of cows giving milk = 150

the total number of cows in corral = 20

Suppose total number of cows giving milk in corral = x

The ratio of the total number of cows present to the total number of cows giving milk should be same in farm and corral

so : 200 = 20

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

150 x

now we do cross multiplication :

200 x = 150* 20

200x = 3000

x= 3000/200

x= 15

SO we expect 15 cows in corral should be giving milk .

Answer : 15

4 0
3 years ago
Read 2 more answers
Please help me ASAP!!
vodomira [7]

It's A.



20characters40characters

4 0
4 years ago
Sine, cosine, and tangent? How would I use them? And they have negatives?​
zhannawk [14.2K]

Answer: Sine - Sin

Tangent - Tan - Cosine - Cos

or inverse Cos-1 - Sin-1 - Tan-1

Step-by-step explanation:

You would use  Sine - Sin for Opposite over hypotenuse

to check it you'd use Sin-1 that is just Sine but inverse that helps you get the answer- I just learned this stuff so bare with me but all of it is the same

I'm hoping I helped give you a small clue  

4 0
3 years ago
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