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3 years ago

The circumference of a circle is 25.12cm. What is the area of the circle? Use 3.14

1 answer:

3 0

C=2pir
area=pir^2

given
c=25.12
25.12=2pir
pi=3.14
25.12=2(3.14)r
25.12=6.28r
divide both sides by 6.25
4=r

sub into other
area=pir^2
area=pi4^2
area=16pi
area=16(3.14)
area=50.24

the area is 50.24 square cm

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Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago

HELP ASAP Three generous friends, each with some cash, redistribute their money as follows: Ami gives enough money to Jan and To

DENIUS [597]

Answer:

$252

Step-by-step explanation:

This is quite a neat question, with no fixed equation. Given that each person gives the other two enough money to double their cash, if Toy had 36 dollars beginning, and 36 at the end - presumably the cash of each person, ( their starting and original ) should be the same as well. Respectively each should be a multiple of 36 dollars.

Jan's " give away " = Ami + 36, Jan - 108, Toy + 72

Toy's " give away " = Ami + 72, Jan + 36, Toy - 108

Therefore, we can conclude that Ami = 144 at the start, presuming he gave away 108 dollars, with a remaining 36. Jan, having 144 dollars ( after having his 72 dollars doubled by Ami ) gives 36 to Ami to double his amount, and 72 to double Toy's doubled amount, remaining with 36 dollars. Now Ami has 72 dollars, Jan has 36, and Toy has 144. Then, Toy double's Ami and Jan's amount, giving away 72 and 36 dollars, remaining with 36 dollars himself. Therefore, Ami has 144 dollars, Jan has 72 dollars, and Toy has 36 dollars both at the beginning and end.

144 + 72 + 36 = 252 dollars ( in total )

6 0

3 years ago

Which expression is equivalent to 4/9(2n - 3)

krek1111 [17]

Answer:

8n/9 - 4/3

Step-by-step explanation:

4 0

3 years ago

J3herjjeodbehi4vdue duehduehd7ehee

Pie

The account already has $278

If you add and then subract 278 then the value will be the same

Your answer is B

Hopw this helps

-GoldenWolfX✔

7 0

3 years ago

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How many "words" can be written using exactly five A's and no more than three B's (and no other letters)?

kakasveta [241]

This is a problem of Permutations. We have 3 cases depending on the number of B's. Since no more than three B's can be used we can use either one, two or three B's at a time.

Case 1: Five A's and One B

Total number of letters = 6

Total number of words possible = $\frac{6!}{5!*1!}=6$

Case 2: Five A's and Two B's

Total number of letters = 7

Total number of words possible = $\frac{7!}{5!*2!}=21$

Case 3: Five A's and Three B's

Total number of letters = 8

Total number of words possible = $\frac{8!}{5!*3!}=56$

Total number of possible words will be the sum of all three cases.

Therefore, the total number of words that can be written using exactly five A's and no more than three B's (and no other letters) are 6 + 21 + 56 = 83

8 0

3 years ago

Read 2 more answers