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mel-nik [20]
3 years ago
14

Find the area of a regular pentagon with a side of 10cm and an apothem of 6.9cm

Mathematics
1 answer:
svetlana [45]3 years ago
7 0
Area = 172.05 i think
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What is [(7x^3)(y^2)]^2/7 in radical form?
GuDViN [60]
Remember
x^ \frac{m}{n} = \sqrt[n]{x^m}
and
(zy)^m=(z^m)(y^m)
and
(x^m)^n=x^(mn)

so
((7x^3)(y^2))^ \frac{2}{7} = \sqrt[7]{((7x^3)(y^2))^2}=
\sqrt[7]{((7x^3)^2(y^2)^2)}=
\sqrt[7]{((7^2x^6)(y^4))}=
\sqrt[7]{((49x^6)(y^4))}=
\sqrt[7]{49x^6y^4}
8 0
3 years ago
(38x-8)(2x+38-6)(8x-10)
Svet_ta [14]

Answer:

608x^3+8840x^2-14048x+2560

Step-by-step explanation:

3 0
3 years ago
What is x+3y=6 what is the x
Ymorist [56]
X=-3y+6 That is the answer.
3 0
3 years ago
Find the domain and the range of the relation. Determine whether the relation is a function.
andre [41]

Answer:

Domain: {-6, -1, 7}

Range: {-9, 0, 9}

The relation is not a function.

Step-by-step explanation:

Given the relation: t{(−1,0),(7,0),(−1,9),(−6,−9)}

In the ordered pairs:

  • The domain is the set of all "x" values
  • The range is set of all "y" values
  • We do not need to list any repeated value in the range/domain more than once.

Domain: {-6, -1, 7}

Range: {-9, 0, 9}

Next, we determine whether the relation is a function.

For a relation to be a function, each x must correspond with only one y value.

However, as is observed in the mapping attached below:

  • f(-1)=0
  • f(-1)=9

The x-value (-1) corresponds to two y-values (0 and 9)

Therefore, the relation is not a function.

8 0
3 years ago
Fine the average of these numbers, 8,12,13,15
GarryVolchara [31]

Answer:

12

Step-by-step explanation:

The average is calculated as

average = \frac{sum}{count} = \frac{8+12+13+15}{4} = \frac{48}{4} = 12

8 0
3 years ago
Read 2 more answers
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