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3 years ago

Find the circumference of the circle. Round your answer to the nearest hundredth. Use 3.14 or pi circle is 3 ft in radius

Mathematics

2 answers:

densk [106]3 years ago

8 0

Answer:

18.84 feet

Step-by-step explanation:

The circumference of a circle can be found using:

c=π*d

The radius is given, but we have to find diameter. The diameter is twice the radius, or

d=2r

The radius is 3 feet. Substitute 3 in for r.

d=2*3

d=6

The diameter is 6 feet.

Now we have the diameter. We can substitute 6 in for d in the circumference formula. We can also substitute 3.14 in for pi.

c=π*d

c=3.14*6

c=18.84

The circumference is 18.84 feet

zalisa [80]3 years ago

5 0

Answer:

18.85ft

Step-by-step explanation:

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Elena L [17]

Answer:

The answer I got for this equation is x=5

I hope this image helps of showing you how to solve the equation.

5 0

3 years ago

A chord has ___________ endpoint(s) on a circle. A. Three B. None C. One D. Two

AveGali [126]

Answer:

The answer is D - Two

Step-by-step explanation:

a chord has two endpoints on a circle

5 0

3 years ago

Given that lines b and c are parallel, and the m<1=4x-30 and m<5=2x+50. Find the measure of angle 1 and measure of angle 7

AfilCa [17]

4x-30=2x+50
+30 +30

4x=2x+80
-2x -2x

2x=80
/2 /2

X=40

4(40)-30=130°

2(40)+50=130°

180-130=50°

Angle 7=50°

5 0

2 years ago

Rationalise the denominator of: 1/(√3 + √5 - √2)

Paul [167]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} }$

can be re-arranged as

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} - \sqrt{2} + \sqrt{5} }$

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} } \times \dfrac{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ {( \sqrt{3} - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{3 + 2 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{5 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{ - ( - \sqrt{3} + \sqrt{2} + \sqrt{5}) }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}}$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}} \times \dfrac{ \sqrt{6} }{ \sqrt{6} }$

$\rm \: = \: \dfrac{- \sqrt{18} + \sqrt{12} + \sqrt{30}}{2 \times 6}$

$\rm \: = \: \dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}$

$\rm \: = \: \dfrac{- 3\sqrt{2} + 2 \sqrt{3} + \sqrt{30}}{12}$

Hence,

$\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} } =\dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}}}$

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<h3>More Identities to know:</h3>

$\purple{\boxed{\tt{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\purple{\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

6 0

3 years ago

Let f(x) = |x| for all real numbers x. Write the formula for the function represented by the described

Ostrovityanka [42]

Answer:

$y=(\frac{1}{3}f(x-3))-1$

Step-by-step explanation:

Please see the picture below.

1. Given the function f(x) = |x|, applying a vertical stretch with scale factor $\frac{1}{3}$ , we have the transformed function:

$y=\frac{1}{3}f(x)$

2. Applying a translation of 3 units to the right, we have:

$y=\frac{1}{3}f(x-3)$

3. Finally applying a translation down of 1 unit, we have:

$y=(\frac{1}{3}f(x-3))-1$

6 0

3 years ago