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kati45 [8]
3 years ago
12

1. Mary used repeated addition to show 3 x 2/5. What would the equation look like?

Mathematics
1 answer:
Hunter-Best [27]3 years ago
6 0

Answer:

3/5 x 2/5

Step-by-step explanation:

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The difference of two numbers is 120 and thier product is 3600 find them?
maksim [4K]

Answer:

(144.8100,24.8128) (x,y)

7 0
3 years ago
Evaluate the summation of 3 times negative 2 to the n minus 1 power, from n equals 1 to 5. A) −93. B) −33. C) 33. D) 93
lions [1.4K]
∑ ( from n = 1 to n =5 ) 3 · ( -2 ) ^( n -1 )
a 1 = 3 · ( -2 ) ^0 = 3
a 2 = 3 · ( - 2 ) = - 6
a 3 = 3 · 4 = 12
a 4 = 3 · ( - 8 ) = - 24
a 5 = 3 · 16 = 48
3 - 6 + 12 - 24 + 48 = 33
Answer: C ) 33
4 0
2 years ago
Read 2 more answers
Distributive property<br> 3(2x – y)
Dima020 [189]

Answer:

6x-3y

Step-by-step explanation:

3 * 2x = 6x

3 * - y = 3y

So 6x - 3y

5 0
3 years ago
Pyramid A has a triangular base where each side measures 4 units and a volume of 36 cubic units. Pyramid B has the same height,
omeli [17]

Answer:

The volume of pyramid B is 81 cubic units

Step-by-step explanation:

Given

<u>Pyramid A</u>

s = 4 -- base sides

V = 36 -- Volume

<u>Pyramid B</u>

s = 6 --- base sides

Required

Determine the volume of pyramid B <em>[Missing from the question]</em>

From the question, we understand that both pyramids are equilateral triangular pyramids.

The volume is calculated as:

V = \frac{1}{3} * B * h

Where B represents the area of the base equilateral triangle, and it is calculated as:

B = \frac{1}{2} * s^2 * sin(60)

Where s represents the side lengths

First, we calculate the height of pyramid A

For Pyramid A, the base area is:

B = \frac{1}{2} * s^2 * sin(60)

B = \frac{1}{2} * 4^2 * \frac{\sqrt 3}{2}

B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}

B = 4\sqrt 3

The height is calculated from:

V = \frac{1}{3} * B * h

This gives:

36 = \frac{1}{3} * 4\sqrt 3 * h

Make h the subject

h = \frac{3 * 36}{4\sqrt 3}

h = \frac{3 * 9}{\sqrt 3}

h = \frac{27}{\sqrt 3}

To calculate the volume of pyramid B, we make use of:

V = \frac{1}{3} * B * h

Since the heights of both pyramids are the same, we can make use of:

h = \frac{27}{\sqrt 3}

The base area B, is then calculated as:

B = \frac{1}{2} * s^2 * sin(60)

Where

s = 6

So:

B = \frac{1}{2} * 6^2 * sin(60)

B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}

B = 9\sqrt 3

So:

V = \frac{1}{3} * B * h

Where

B = 9\sqrt 3 and h = \frac{27}{\sqrt 3}

V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}

V = \frac{1}{3} * 9 * 27

V = 81

6 0
3 years ago
Read 2 more answers
Quick what’s the answer
GaryK [48]
Use the Pythagorean Theorem of a² + b² = c² to solve for x. 

x² + (x + 3)² = (√117)²
x² + x² + 6x + 9 = 117
2x² + 6x + 9 = 117
2x² + 6x + 9 - 117 = 0
2x² + 6x - 108 = 0
2x² + 18x - 12x - 108 = 0
2x(x + 9) - 12(x + 9) = 0
(2x - 12)(x + 9) = 0

x = - 9, 6

Length cannot be negative so you can't use - 9. 

x = 6. Option C is your answer. 
6 0
2 years ago
Read 2 more answers
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