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steposvetlana [31]
2 years ago
13

Solve for E: E=〖mc〗^2 m = 3 c = 6

Mathematics
2 answers:
GenaCL600 [577]2 years ago
7 0
The answer for the problem is E=324
rewona [7]2 years ago
4 0

Answer:

E  =  324

Step by step explanation:

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GREYUIT [131]
Do you have any formula ?
5 0
3 years ago
Pizzeria Napoli sells a round pizza with diameter 16 inches and a square pizza with side length 15 inches. Which of the two shap
zavuch27 [327]

Answer: Square, 24 square inches

Step-by-step explanation:

Given

Diameter of round pizza d=16\ in.

Side of square pizza a=15\ in.

Area of circle \pi r^2

The area of a square is a^2

\therefore \text{Area of round Pizza}=\pi \times (\frac{16}{2})^2\\\Rightarrow A_1=\pi \times 8^2=201.088\ in.^2

The area of square pizza is

A_2=15^2=225\ in.^2

Clearly, the area of square pizza is more and the difference in their area is

\Rightarrow A_2-A_1=225-201.088=23.91\approx\ 24\ in.^2

6 0
2 years ago
The value of w???????
goldenfox [79]

Answer:

Step-by-step explanation:

The angle bisector of a triangle divides the opposite side in 2 segments that are proportional to the other 2 sides of the triangle. Namely:

\frac{15}{18}=\frac{w}{30} and cross multiply.

18w = 450 so

w = 25

4 0
3 years ago
A store has 80 modems in its inventory, 30 coming from source a and the remainder from source
Mrrafil [7]
The number of defective modems in the inventory is 20%⋅ 30 + 8%⋅ 50 = 10 (out of 80). Note that the number of defectives in the inventory is fixed, i.e., we are not told that there is 1 8 probability that a modem in the inventory is defective, but rather that exactly 1 8 of all modems are defective. The probability that exactly two modems in a random sample of five are defective is = 0.102
5 0
3 years ago
Please need help evaluate the math problems
alina1380 [7]

x + y when x = 6 and y = 2

6 + 2 = 8

\frac{9x+24}{4} when x = 12 and y = 20

\frac{148}{4} = 37

\frac{6m}{n} when m = 12 and n = 3

\frac{72}{3} = 24

Six times x

6x

Six increased by x

6 + x

The sum of x and y

x + y

6 0
3 years ago
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