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Temka [501]
4 years ago
6

8,11, and 14 please. Thank you

Mathematics
2 answers:
Minchanka [31]4 years ago
8 0
Formula: A= 3.14xr^2
8. 39x39 = 1,521x3.14 = 4,775.94

11. 29x29 = 841x3.14 = 2,640.74

14. 22.5x22.5 = 506.25x3.14 = 1,589.6

I hope this helps
stellarik [79]4 years ago
6 0
8= 266.9
11= 91.06
14= 70.65
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The following table is a function.<br> х -1 5 7 2 6 3 8 4 1 y -3 3 7 -4 5 9 7 1 0
Kruka [31]

Answer:false

Step-by-step explanation:

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Select all the pairs of equivalent expressions.
iren2701 [21]

Answer:

\huge\boxed{(3^4)^4=3^8\cdot3^8}\\\boxed{4^3\cdot5^3=20^3}\\\boxed{(4^3)^3=4^3\cdot4^3\cdo4^3}

Step-by-step explanation:

a^n\cdot a^n=a^{n+m}\\\\(a^n)^m=a^{n\cdot m}\\\\(a\cdot b)^n=a^n\cdot b^n\\=====================

(4^3)^3=4^{3\cdot3}=4^9\\4^3\cdot4^3=4^{3+3}=4^6\\\\4^9\neq4^6\to(4^3)^3\neq4^3\cdot4^3\\==================

(3^4)^4=3^{4\cdot4}=3^{16}\\3^8\cdot3^8=3^{8+8}=3^{16}\\\\(3^4)^4=3^8\cdot3^8\\================

6^4\cdot3^4=(6\cdot3)^4=18^4\\\\18^4\neq18^8\to6^4\cdot3^4\neq18^8\\=================

4^3\cdot5^3=(4\cdot5)^3=20^3\\\\4^3\cdot5^3=20^3\\=================

(4^3)^3=4^{3\cdot3}=4^9\\4^3\cdot4^3\cdot4^3=4^{3+3+3}=4^9\\\\(4^3)^3=4^3\cdot4^3\cdot4^3

3 0
3 years ago
Hey Bestie! 50 pts Pls help! Real answers only pls &lt;3
frosja888 [35]

Answer:

1. \: 3i

2. option D

3. option C

4. option D

5. option C

6. option B

7. option C

8. option D

9. option C

10. option C

Step-by-step explanation:

<h2>1. \:  \sqrt{ - 9}</h2>

\sqrt{ - 1(9)}

\sqrt{ - 1}  \times  \sqrt{  9}

i \times  \sqrt{9}

i \times  \sqrt{ {3}^{2} }

i \times 3

3i

<h2>2. \:  \sqrt{ - 8}</h2>

\sqrt{ - 1(8)}

\sqrt{ - 1}  \times  \sqrt{8}

i \times  \sqrt{8}

i \times  \sqrt{ {2}^{2} \times 2 }

2i \sqrt{2}

<h2>3. \:  \sqrt{ - 80}</h2>

\sqrt{ - 1}  \times  \sqrt{80}

i \times  \sqrt{80}

4i \:  \sqrt{5}

<h2>4. \:  \sqrt{ - 75}</h2>

\sqrt{ - 1}  \times   \sqrt{75}

i \times  \sqrt{75}

i \times  \sqrt{ {5}^{2} \times 3 }

5i \sqrt{3}

<h2>5. \:  \sqrt{ - 72}</h2>

\sqrt{ - 1}  \times  \sqrt{72}

i \times  \sqrt{72}

i \times ( {6}^{2}  \times 2)

6i \sqrt{2}

<h2>6.  \sqrt{ - 20}</h2>

\sqrt{ - 1}  \times  \sqrt{20}

i \times  \sqrt{20}

i \times  \sqrt{ {2}^{2}  \times 5}

2i \sqrt{5}

<h2>7. \:  \sqrt{ - 27}</h2>

\sqrt{ - 1}  \times  \sqrt{27}

i \times  \sqrt{27}

i \times  \sqrt{ {3}^{2}  \times 3}

3i \sqrt{3}

<h2>8. \:  \sqrt{ - 12}</h2>

\sqrt{ - 1 \times 12}

i \times  \sqrt{12}

i \times  \sqrt{4(3)}

2i \sqrt{3}

<h2>9. \:  \sqrt{ - 125}</h2>

\sqrt{ - 1}  \times  \sqrt{125}

i \times  \sqrt{ {5}^{2} \times 5 }

5i \sqrt{5}

<h2>10. \:  \sqrt{ - 180}</h2>

\sqrt{ - 1}  \times  \sqrt{180}

i \times  \sqrt{ {6}^{2} \times 5 }

6i \sqrt{5}

<h3>Hope it is helpful...</h3>
5 0
3 years ago
Jamal uses the steps below to solve the equation 6x – 4 = 8.
LenaWriter [7]
B.)  because according to the identity property of addition, any number added to zero will remain the same.  
6 0
3 years ago
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1. Find the Least Common Multiple of these two monomials:<br> See picture
gtnhenbr [62]

Answer:

<em>The last choice is correct</em>

<em />LCM=120a^4b^7c^5<em />

Step-by-step explanation:

<u>Least Common Multiple (LCM)</u>

To find the LCM we can follow this procedure:

List the prime factors of each monomial.

Multiply each factor the greatest number of times it occurs in either factor.

We have two monomials:

12a^4b^2c^5

40a^3b^7c^1

The prime factors of the first monomial are:

2^2,3,a^4,b^2,c^5

The prime factors of the second monomial are:

2^3,5,a^3b^7c^1

LCM = Multiply 2^3*3*5*a^4*b^7*c^5

These are all the factors the greatest number of times they occur.

Operating:

LCM=8*15*a^4*b^7*c^5

\boxed{LCM=120a^4b^7c^5}

The last choice is correct

3 0
4 years ago
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