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Katyanochek1 [597]
3 years ago
10

what is the arc length of a circle that has a 6-inch radius and a central angle that is 65 degrees? use 3.14 for π and round you

r answer to the nearest hundredth. a.) 0.65 inch b.) 1.13 inches c.) 6.80 inches d.) 390.01 inches
Mathematics
2 answers:
shepuryov [24]3 years ago
7 0

Answer:

Option c is correct

l = 6.80 inches

Step-by-step explanation:

The arc length(l) of circle is given by:

l = r \theta                     .......[1]

where,

r is the radius of the circle and

\theta is the central angle in radian.

As per the statement:

radius(r) = 6 inch

\theta = 65^{\circ}

Use conversion:

1 degree = π/180=\frac{3.14}{180}  radian

then;

65 degree = 1.13388889 radian.

Substitute the given values in [1] we have;

l = 6 \cdot 1.13388889

⇒l = 6.80333334 inches

Therefore, the arc length of a circle to the nearest hundredth is,  6.80 inches

Vinvika [58]3 years ago
3 0
The answer is C.

The arc length is 6.806784.

Hope this helps.
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What is that simplest form of 9*5/6?​
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Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.
Mice21 [21]

Answer:

<em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

Step-by-step explanation:

Making factors of 8y^{6}+ 144y^{5}+ 640y^{4}

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\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)

Using <em>factorization</em> method:

\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times  4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)

Now, Making factors of 2y^{4} + 40y^{3} + 200y^{2}

Taking 2y^{2} common:

\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)

Using <em>factorization</em> method:

\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10)        ............ (2)

The underlined parts show the Highest Common Factor(HCF).

i.e. <em>HCF</em> is 2y^{2} (y+ 10).

We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

HCF \times LCM = p \times q

Using equations <em>(1)</em> and <em>(2)</em>: \Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)

Hence, <em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

5 0
3 years ago
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vaieri [72.5K]

Answer:

The three numbers are 341, 342, and 343

Step-by-step explanation:

We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 1026. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 1026

To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:

X + X + 1 + X + 2 = 1026

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Which means that the first number is 341, the second number is 341 + 1 and the third number is 341 + 2. Therefore, three consecutive integers that add up to 1026 are 341, 342, and 343.

341 + 342 + 343 = 1026

We know our answer is correct because 341 + 342 + 343 equals 1026 as displayed above.

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