Identify the 19th term of a geometric sequence where a1 = 14 and a9 = 358.80. Round the common ratio and 19th term to the neares

Response: 3.2

Detailed explanation:

Answer: C) HF measures 4 units and GH is 2 units.

Step-by-step explanation:

The SSS similarity theorem asserts that triangles are similar if the lengths of their corresponding sides are in proportion.

For triangles ΔDFE and ΔGFH:

DG equals 15, GF is 5, EH equals 12, and DE is 8.

To demonstrate the similarity of ΔDFE and ΔGFH according to the SSS similarity theorem, we require:

Thus, to confirm that △DFE is similar to △GFH utilizing the SSS similarity theorem and the data from the diagram, it is essential to establish that HF measures 4 units and GH measures 2 units.

Answer: Initially, he had 27 sweets.

Step-by-step explanation: The most logical approach is to work backwards from what remained after the third day to the start of the first day.

On the third day, he consumed one-third of his sweets and was left with 8. If we let the total sweets on day three be denoted as a, then one-third of a equals what he ate and the two-thirds left equals 8, giving us:

8/a = 2/3

By cross-multiplying, we find:

8 x 3 = 2a

Therefore, 24 = 2a

This leads to a = 12.

Let the sweets on day two be represented as b. If he consumed one-third of b and was left with 12, we have the same structure; hence:

12/b = 2/3

Cross-multiplying gives:

12 x 3 = 2b

So, 36 = 2b, leading to b = 18.

Denote the number of sweets on day one as x. If one-third of x was eaten and 18 remained, we can set up the equation:

18/x = 2/3

Again, cross-multiplying results in:

18 x 3 = 2x

Which simplifies to 54 = 2x, yielding x = 27.

Thus, Tim received 27 sweets at the start.

The pot has the capability to hold 1/3 more than its current level. Therefore, 1/3 of 5 1/2 quarts equals:

(1/3)(11/2) = 11/6 quarts. Thus, the total capacity of the pot is 11/6 quarts.

Step-by-step explanation: Start by graphing the constraints represented by the lines

5x + y = 16, which has intercepts at (16, 0) and (0, 16), and

The vertices of the feasible region are determined to be (0, 0), (0, 16), (2, 6), and (11, 0). Evaluate the objective function at each vertex.

At (0, 0), P = 0. At (0, 16), P = 32. At (2, 6), P = 18. Finally, at (11, 0), P = 33. Thus, the maximum P value occurs when x = 11 and y = 0.