40º
7) In this problem, we can see that both tangent lines to that circle come from the same point O.
So, we can write out the following considering that there is one secant line DO and one tangent line to the circle AO

A(n) = a₁.(r)ⁿ⁻¹, where a₁ = 1st term, r= common ratio and n, the rank
In the formula given a₁ = 5, r = 3/2 and n = 6 (we have to find the 6th term value).
a₆ = 5.(3/2)⁶⁻¹ = 5.(3/2)⁵ = 1215/32 (answer C)
Answer:
1529.4 m^3
Step-by-step explanation:
We will use the formula V = 1/3*h*B for the volume of the pyramid, where V is volume, h is height, and B is are of the base.
First, since we know h and are looking for V, we need B. To find this, since it has a square base, we multiply b, the base length, by itself. 15.3m * 15.3m = 234.09m^2.
Next, we plug into the formula.
V = 1/3*19.6m*234.09m^2
And simplify:
V= 1/3 * 4,588.164m^3
V= 15.29.3879....
Last, round to the tenth
1529.4 m^3
Remainder Theorem: when a polynomial <em>P(x) </em>is divided <em>by (x - a), </em>the remainder is <em>P(a).</em>
<em />
Given expression:

In this case:


To know the reminder, we have to plug in the value of <em>z </em>in the polynomial:

Simplifying:



Answer: The remainder is 15.