This whole problem is only about a circle. Let's not get confused by the rest of the lampshade.
The circumference of any circle is (pi) times (the diameter) .
For this circle, it's (pi) times (13 cm).
They told us what to use for (pi), so it's (3.14) x (13 cm) = <u>40.82 cm</u>
and that's all there is to it.
Given that , Length of Base and Hypotenuse is 12 cm and 24 cm , respectively .
Need To Find : The measure of Angle x .
<em>As </em><em>We </em><em>know </em><em>that </em>,
Cos ∅ = Base / Hypotenuse
ㅤㅤ <u>By </u><u>Substituting </u><u>known </u><u>Values </u><u>:</u>
ㅤ⤀Cos x = 12/24
ㅤ⤀Cos x = ½
ㅤ⤀Cos 60⁰ = ½
Therefore,
Step-by-step explanation:
1) Your problem → (4x^2 - 17x^3 + 9) - (x^2 + 9x + 23x^2 + 11)
(-17x^3+4x^2+9)-(x^2+23x^2+9x+11)
=-17x^3+4x^2+9-x^2-23x^2-9x-11
=-17x^3+4x^2-x^2-23x^2-9x+9-11
=-17x^3-20x^2-9x-2
2) Your problem → 0 - 19.73 - 25x^2 - 12x - 3
=0-19.73-25x2-12x-3
=-25x^2-12x-22.73
3) - 10.x^3 – 162x^2 – 24x - 4
4) Your problem → 17x^3 - 20x^2 - 9x^2
=17x^3-20x^2-9x^2
=17x^3-29x^2
5) -16x^3 – 243x^2 – 12x – 3
Answer:
pls send a full pic of the picture
Answer:
52
Step-by-step explanation:
A cubical box without a top is 4 cm on each edge containing 64 identical 1 cm cubes.
We have to find the number of small cubes that actually touch the box.
Now, there are 4 layers of 16 cubes from bottom to top.
In the bottom-most layer, all the 16 small cubes will touch the bottom of the large cube.
Again, from the top three layers, only the outer small cubes will touch the sides of the large cube.
So, there are 12 small cubes in each top three layers that will touch the large cube.
Hence, in total there will be (16 + 12 ×3) = 52 small cubes that actually touch the large cube box. (Answer)