Consider one pyramid
Side length of base = 1.5cm and its height is 1 cm
Slant height of one of the lateral faces = sqrt(1^2 + 0.75^2) = 1.25 cm
Area of one of the triangular faces = 0.5 * 1.5* 1.25 = 0.9375 cm^2
There are 8 of these so the required surface area = 8 * 0.9375
= 7.5 cm^2 Answer
Answer:
11. x = -3+√37 ≈ 3.08276
12. x = 11.2
13. x = -6 +6√5 ≈ 7.41641
Step-by-step explanation:
In each case, the relation of interest is ...
(distance to circle near) × (distance to circle far) = (distance to circle near) × (distance to circle far)
When there is only one point of intersection of the secant with the circle—because it is a tangent—then the product is the square of the length of the tangent.
11. 2(2+12) = x(x +6)
x² +6x -28 = 0
(x +3)² -37 = 0
x = -3+√37 ≈ 3.08276
12. 5(5+x) = 9(9)
5x +25 = 81
x = 56/5 = 11.2
13. x(x +12) = 12(12)
x² +12x -144 = 0
(x +6)² -180 = 0
x = -6 +√180 ≈ 7.41641
_____
<em>Comment on this secant rule</em>
This rule turns out to apply whether the point of intersection of the secant lines is outside the circle (as in these problems) or inside the circle (as in problem 9). The product of the two distances from the point of intersection to the circle is a constant for a given pair of intersecting secants/chords.
A:
add up all the money he deposited
3.98+51.02+38.52+12.70=$106.22
and all the money withdrew:
-5.23-5.22-3.50-4.39=-18.34
b-
3.50, 3.98, 4.39, 5.22, 5.23, 12.7 38.52 51.02
just look at the digits from left to right the bigger they are the bigger the number is.if they were equal look at the next digit:
example:
3.50 and 12.7
there is a 1 in 12.7 tenth place but 3.50 doesn't even have a tenth place(or rather, it has a zero on it's tenth place) so 12.7 is bigger than 3.50
c-add all deposits and subtract all withdrawals:
106.22+98-18.34=185.88
