Answer:
I will do that buddy. I'll have fun
I'm not certain but I think it's <span>4 2/8. </span>
The answer is (-∞.∞).
Draw out a number line. Plot 0 and -4 on the number line.
Shade to the left of x = 0, and have a filled in circle at the endpoint. This is the graph of x≤0
Then graph x ≥−4 by plotting a filled in circle at -4, and shading to the right.
Note how the two graphs overlap to cover the entire real number line
So if we have x≤0 or x≥−4 then we're basically saying x is any real number. To write this in interval notation, we write (−∞,∞)
This is the interval from negative infinity to positive infinity (or just infinity). We exclude each endpoint because we can't actually reach infinity itself. Infinity is not a number. Infinity is a concept
Side note: if you change the "or" to "and", then the solution to x≤0 and x≥−4 would be [-4, 0][−4,0] to indicate the interval from x = -4 to x = 0, including both endpoints. This is the region where the two graphs overlap.
To know more about graphs
visit-:brainly.com/question/10712002
SPJ9
Firstly, find the radius with the formula diameter (d) : 2
8 : 2 = 4 cm
The Vilume formula is r²hπ = 4²*12*π = 16*12*π = 192π cm³
Your exercise says to use "3" as a value for π
Then V is 192*3 = 576 cm³
Answer:
9. (7a + 6b – 9c) – (3a – 6c)
=7a+6b-9c-3a+6c
=7a-3a+6b-9c+6c
=4a+6b-3c
10. (x2 – 9) – (-2x2 + 5x – 3)
= x^2-9+2x^2-5x+3
=x^2+2x^2-5x-9+3
=3x^2-5x-6
11. (5 – 6d – d2) – (-4d – d2)
=5-6d- d^2+4d+ d^2
=5-6d+4d-d^2+ d^2
=5-2d
12. (-4x + 7) – (3x – 7)
=-4x+7-3x+7
= -4x-3x+7+7
=-7x+14
13. (4a – 3b) – (5a – 2b)
=4a-3b-5a+2b
=4a-5a-3b+2b
= -a-b
14. (2c + 3d) – (-6d – 5c)
=2c+3d+6d+5c
=2c+5c+3d+6d
=7c+9d
15. (5x2 + 6x – 9) – (x2 – 3x +7)
=5x^2+6x-9- x^2+3x-7
=5x^2- x^2+6x+3x-9-7
=4x^2+9x-16
16. (3y – 6) – (8 – 9y)
=3y-6-8+9y
=3y+9y-6-8
=12y-14
17. (3a2 – 2ab + 3b2) - (-a2 – 5ab + 3b2)
=3a^2-2ab+3b^2+ a^2+5ab-3b^2
=3a^2+ a^2-2ab+5ab+3b^2- 3b^2
=4a^2+3ab
18. 5c – [8c – (6 – 3c)]
=5c-[8c-6+3c]
=5c-8c+6-3c
=5c-8c-3c+6
= -6c+6
19. 10x + [3x – (5x – 4)]
=10x+[ 3x-5x+4]
=10x+3x-5x+4
=8x+4
20. 3x 2 – [7x- (4x – x2) + 3]
=3x^2-[7x-4x+ x^2+3]
=3x^2-7x+4x- x^2-3
=3x^2-x^2-7x+4x-3
=2x^2-3x-3
21. x2 – [ - 3x+ ( 4 – 7x)]
= x^2-[ -3x+4-7x]
= x^2+3x-4+7x
= x^2+3x+7x-4
= x^2+10x-4
