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mixer [17]
3 years ago
10

What if the perimeter of the table?

Mathematics
2 answers:
Nitella [24]3 years ago
4 0
Ans=>12+8(pi)
If you look closely at those corners u can see those are the quarters of a circle with radius 4.
So now just add the lenghts and the circumference of that circle of radius 4
3+3+3+3+8pi
Vlad1618 [11]3 years ago
3 0

Step-by-step explanation:

13? i forgot all this math XD

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What is the slope-intercept equation of the line below?
mylen [45]
The slope intercept equation of the line above is y=-2x+3
8 0
3 years ago
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5x + 2y = 12<br><br> 5x – 2y = 28<br> Solve by substitution
andrey2020 [161]

Answer:

Divide each term by  

5

and simplify.

Tap for more steps...

x

=

2

y

5

+

12

5

4

x

=

28

−

3

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in  

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3

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with  

2

y

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+

12

5

.

x

=

2

y

5

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4

(

2

y

5

+

12

5

)

=

28

−

3

y

Simplify  

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(

2

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5

+

12

5

)

.

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x

=

2

y

5

+

12

5

8

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5

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48

5

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−

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Move all terms in  

8

y

5

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48

5

=

28

−

3

y

to the left side and simplify.

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x

=

2

y

5

+

12

5

23

(

y

−

4

)

5

=

0

Solve for  

y

in the second equation.

Tap for more steps...

x

=

2

y

5

+

12

5

y

=

4

Replace all occurrences of  

y

in  

x

=

2

y

5

+

12

5

with  

4

.

x

=

2

(

4

)

5

+

12

5

y

=

4

Simplify  

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(

4

)

5

+

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x

=

4

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=

4

The solution to the system of equations can be represented as a point.

(

4

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4

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The result can be shown in multiple forms.

Point Form:

(

4

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4

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Equation Form:

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7 0
3 years ago
Does anyone know how to solve this 5 and 6
Sveta_85 [38]
5. (2,-6) (-2,6)
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7 0
3 years ago
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Call an integer $n$ oddly powerful if there exist positive integers $a$ and $b$, where $b&gt;1$, $b$ is odd, and $a^b = n$. How
azamat

Answer:

There are 16 oddly powerful integers less than 2010

Step-by-step explanation:

∵ b is an odd integer

∵ b > 1

∴ The first value of b is 3

∵ a is an integer

- We can use a = 1, 2, 3, ..........

∵ a^{b}=n

∵ n < 2010

- Let a = 1, 2, ............... 12 because 12³ is greatest integer  < 2010

∵ 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125, 6³ = 216, 7³ = 343,

   8³ = 512, 9³ = 729, 10³ = 1000, 11³ = 1331, 12³ = 1728

∴ There are 12 oddly powerful integers with b = 3

Now the second value of b is 5

1^{5}=1 but we took 1 before so we will start with 2

∵ 2^{5}=32, 3^{5}=243, 4^{5}=1024

- 4^{5} is the greatest integer < 2010

∴ There are 3 oddly powerful integers with b = 5

Now the third value of b is 7

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- 2^{7} is the greatest integer < 2010

∴ There is 1 oddly powerful integers with b = 7

Now the fourth value of b is 9

∵ 2^{9}=512

- 2^{9} is the greatest integer < 2010

- But we used 512 before

∴ There is no oddly powerful integers with b = 9

- 9 is the greatest value of b which makes a^{b}

∵ 12 + 3 + 1 = 16

∴ There are 16 oddly powerful integers less than 2010

5 0
3 years ago
I needd helppp plss and tyyy
ale4655 [162]

Answer:

area : 0.0001539 square kilometers (km²) 153.9 square meters (m²)

Step-by-step explanation:

A circle of radius = 7 or diameter = 14 or circumference = 43.98 meters has an  area of: 0.0001539 square kilometers (km²) 153.9 square meters (m²)

hope that helps

7 0
2 years ago
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