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Pavel [41]
3 years ago
13

6. What is the approximate volume of the cone? Use for a T. 3cm 14 cm

Mathematics
1 answer:
Tomtit [17]3 years ago
6 0
If 3cm is the radius and 14cm is the height, the volume would be 131.95cm^3. In terms of pi it would be 9pi(14/3).
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Round to the nearest hundreds place<br> Show your work<br> 4x = 7 - 8x - 27
faltersainse [42]

Answer:

x = 5

Step-by-step explanation:

4x = 7 - 8x - 27

1) Combine like terms (7 and -27). So, 7 + (-27) = -20.

        4x = 8x - 20

2) Subtract 8x from both sides of the equation. So, 4x - 8x = -4x.

        -4x = -20

3) Divide both sides by -4. So, -20 / -4 = 5.

        x = 5

5 0
3 years ago
Which number line shows the approximate location of 35?
ANEK [815]

Answer:

I need a list of numberlines or some other kind of explaination on how to solve this problem. I caint help you on this with out them. Sorry.

Step-by-step explanation:

5 0
3 years ago
4) There are 4 marbles and 3 cubes in a board game. The marbles are black, blue, yellow, and red. The cubes are numbered 1, 2, a
krek1111 [17]

Answer:

<u>Here is the sample space, its size is 4*3 = 12:</u>

  • Black 1
  • Black 2
  • Black 3
  • Blue 1
  • Blue 2
  • Blue 3
  • Yellow 1
  • Yellow 2
  • Yellow 3
  • Red 1
  • Red 2
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3 0
3 years ago
Read 2 more answers
FIND THE SUM AND THE DIFFERENCE FOR EACH PAIR OF POLYNOMIALS please show work
Pepsi [2]

Answer:

12a. Addition = 7y⁵ + 7y³ – 6y – 5

12b. Subtraction = – y⁵ – 3y³ + 8y + 11

13a. Addition = 5x⁴ + x³ – 6x² + 8x + 17

13b. Subtraction = – x⁴ – 9x³ –6x² + 8x + 3

14a. Addition = – 2b⁴ + b³ – 11b² – 1

14b. Subtraction = 4b⁴ + 7b³ – b² + 8b – 1

15a. Addition = m⁵ + m⁴ + m³ + m²

15b. Subtraction = m⁵ – m⁴ – m³ – m² – 10

16a. Addition = 4x⁴ + 12x³ + 18x² + 16x + 4

16b. Subtraction = 4x⁴ – 4x³ + 14x² – 16x + 4

Step-by-step explanation:

12a. Addition

.. 3y⁵ + 2y³ + y + 3

+ (4y⁵ + 5y³– 7y – 8)

————————————

= 7y⁵ + 7y³ – 6y – 5

12b. Subtraction

.. 3y⁵ + 2y³ + y + 3

– (4y⁵ + 5y³– 7y – 8)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – y⁵ – 3y³ + 8y + 11

13a. Addition

.. 2x⁴ – 4x³ – 6x² + 8x + 10

+ (3x⁴ + 5x³ +...................+ 7)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 5x⁴ + x³ – 6x² + 8x + 17

13b. Subtraction

.. 2x⁴ – 4x³ – 6x² + 8x + 10

– (3x⁴ + 5x³ +...................+ 7)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – x⁴ – 9x³ – 6x² + 8x + 3

14a. Addition

...... b⁴ + 4b³ – 6b² + 4b – 1

+ (– 3b⁴ – 3b³ – 5b² – 4b)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – 2b⁴ + b³ – 11b² – 1

14b. Subtraction

...... b⁴ + 4b³ – 6b² + 4b – 1

– (– 3b⁴ – 3b³ – 5b² – 4b)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4b⁴ + 7b³ – b² + 8b – 1

15a. Addition

... m⁵ + m³ – 5

+ (m⁴ + m² + 5)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= m⁵ + m⁴ + m³ + m²

15b. Subtraction

... m⁵ + m³ – 5

– (m⁴ + m² + 5)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= m⁵ – m⁴ – m³ – m² – 10

16a. Addition

.... 4x⁴ + 8x³ + 16x² + 4

+ (4x³ + 2x² + 16x)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4x⁴ + 12x³ + 18x² + 16x + 4

16b. Subtraction

.... 4x⁴ + 8x³ + 16x² + 4

– (4x³ + 2x² + 16x)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4x⁴ – 4x³ + 14x² – 16x + 4

3 0
3 years ago
Which equation can be used to represent six added to twice the sum of a numbee and four is equal to one half of the difference o
Ronch [10]

Answer:  6 + 2 (x+4) =   \frac{3}{2}(3-x)

Step-by-step explanation:

let the number be x , then :

six added to twice the sum of a number and four means :

6 + 2 (x+4)

one half of the difference of three and the number means :

\frac{3}{2}(3-x)

combining the two , we have

6 + 2 (x+4) = \frac{3}{2}(3-x)

6 0
4 years ago
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