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ryzh [129]
3 years ago
8

Please answer this by midnight I need it for Algebra 3 and I don't want to fail please

Mathematics
1 answer:
Monica [59]3 years ago
6 0

The length of the arc is about 1.758 cm.

First, we need to determine the size of the angle. If we have pi/5 radians, we can convert that to 36 degrees, because pi radians is 180 degrees. Now, we know the measure of the arc is 36 degrees. 36 degrees out of 360 degrees is 10%.

Our arc is 10% of the circumference of the circle.

The circumference is 2(pi)r or 2(3.14)2.8 = 17.58 cm

Now, multiply 17.58 by 0.1 to get a total of 1.758 cm.

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What is 5,254,071,926 written in standard form
Elis [28]
5.254071926×10^9
Standard form is taking a large number and putting it in a form where the number is smaller than 10. To do that (using the number above) you would take the first number which is 5 and then make every number after it part of a decimal. You then use [×10^x] (multiplied by 10 to the power of the amount of numbers after the decimal point (In this case 9)) to show others that to find the full number you multiply the decimal by 10 to the power of 9.
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4 years ago
( please help me) ( 11 points)
lara [203]

Answer:

500cm2

Step-by-step explanation:

2(9×20)+(7×20)

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5 0
3 years ago
The hawaiian alphabet has 12 letters. How many permutations are possible for five of these letters
guajiro [1.7K]

Answer: The answer is 50980

Step by Step Explanation:

First, temporarily assume that two letters with I are different, call them i 1 and i 2. Three "a" are also called as a 1, a 2 and a 3, and two h as h 1 and h 2. Then there are 11 * 10 * 9 * 8 * 7 = 55440 possible "words" (one of 11 is the first letter, 10 is the second, and so on). But because equal letters do the same "words," some "words" were counted twice or more. We have to deduct the number of "parasitic" counts although it is fairly small. The words that counted more than once are divided into many disjoint sets: 1) with two I but without repetitions of a and h; 2) with two h but without repetitions of a and I 3) with two a but without repetitions of I and h; 4) with three a but without repetitions of I and h; 5) with two I and two a's; 6) two i's and tree a's; 7) two i's and two h's 8) two h's and two a's; 9) two h's and one tree. The first category includes terms counted twice and its scale is (5 * 4) * (6 * 5 * 4) = 2400 (the first I stays at one of the 5 positions, the second at one of the 4, then 11-2i-1h-2a=6). So we have 2400/2 = 1200 to subtract. Group 2 gives -600 as well, and group 3 also. Group 4 gives * (6 * 5) = 1800 (5 * 4 * 3), and the terms are counted 6 times, -300. Groups 5, 7 , 8: 5 * 4 * 3 * 2 * 6 = 720 and counted four times, therefore -180. Group 6 and 9: 5 * 4 * 3 * 2 * 1 = 120, with 12 counts, -10. Altogether -(1200 * 3 + 300 + 180 * 3 + 10 * 2) = -4460.The answer will be 55440-4460 = 50980.

3 0
4 years ago
A. Show
nydimaria [60]

a. Recall the double angle identities:

\sin^2x=\dfrac{1-\cos2x}2

\cos^2x=\dfrac{1+\cos2x}2

Then

\sin^2x\cos^2x=\dfrac{(1-\cos2x)(1+\cos2x)}4=\dfrac{1-\cos^22x}4

Applying the identity again, we have

\sin^2x\cos^2x=\dfrac{1-\frac{1+\cos4x}2}4=\dfrac{2-(1+\cos4x)}8=\dfrac{1-\cos4x}8

as required.

b. Using the result from part (a),

\sin^2x\cos^2x=\dfrac{1-\cos4x}8=\dfrac{2-\sqrt2}{16}

\implies\cos4x=\dfrac1{\sqrt2}

\implies4x=\dfrac\pi4+2n\pi\text{ or }4x=-\dfrac\pi4+2n\pi

(where n is any integer)

\implies\boxed{x=\pm\dfrac\pi{16}+\dfrac{n\pi}2}

8 0
4 years ago
Express √-225 in its simplest terms
Montano1993 [528]

Answer:

Step-by-step explanation:

4 0
3 years ago
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