If AE=6x-55 and EC=3x-16, find DB. (Hint: Find x first and then substitute.)

Complete question:

Benjamin treats himself to breakfast at his go-to restaurant. He orders chocolate milk priced at \$3.25$3.25dollar sign, 3, point, 25. Next, he aims to purchase as many pancake stacks as possible while keeping his total at or below \$30$30dollar sign, 30 prior to tax. Pancakes are sold in stacks of 4 at \$5.50$5.50dollar sign, 5, point, 50. Let SSS denote the number of pancake stacks purchased by Benjamin. 1) What inequality represents this situation?

Answer:

Refer to the explanation below.

Step-by-step explanation:

Information provided:

Chocolate milk costs = $3.25

Price of pancake stack = $5.50 (for 4 pancakes)

Pancake stacks bought = S

Maximum spending ≤ $30

Chocolate milk cost + (Cost per pancake stack × number of stacks) ≤ $30

3.25 + 5.50S ≤ 30

5.50S ≤ 30 - 3.25

5.50S ≤ 26.75

S ≤ 26.75 / 5.50

S ≤ 4.86

Therefore, the maximum number of pancake stacks he can buy without going over budget is 4.

Thus, total pancakes = stacks × pancakes per stack

= 4 × 4

= 16

Answer:

E(t) = [ 4cos(t), 2 sin(t) ]

Step-by-step explanation:

The equation for the ellipse can be represented as:

 (x² ÷ a² ) + (y² ÷ b² ) = 1    Here, a and b symbolize the semi-major and semi-minor axes.

In this specific instance, we have    x²/16 + y²/ 4 = 1

This expression indicates that it follows the form of (x/a)² + (y/b)² =1

Specifically in our case,    x²/16  + y²/4 = 1, which resembles                 sin²α + cos²α = 1.

By setting x = 4 cos(t), we can proceed to calculate y.

Utilizing the equation x²/16  + y²/4 = 1, we find that:

x²/16  + y²/4 = 1   ⇒    (x²  +  4y²) ÷ 16  = 1

Solving for y yields: (x²  +  4y²)  = 16    ⇒  y²  = ( 16 - x²) ÷4

Substituting x = 4 cos(t) gives us y²  =  (16 - 16cos²(t)) ÷ 4, leading to y = √4 (1 - cos²(t)).

Consequently, we have y = 2 sin(t).

Thus, the vector parameterization of the ellipse can be stated as:

E(t) = [ 4cos(t), 2 sin(t) ]

Answer: After the withdrawal, Rajan's bank balance will be

6150 RS - 2050 RS = 4100 RS

Step-by-step explanation:

Initially, Rajan had 6000 RS in his bank account about a week ago. This indicates that the initial balance he possessed is 6000 RS.

One week later, he credited 150p RS to his account.

Thus, the current total in his account is

= 6000 RS + 150p RS = 6150 RS.

The following day, he withdrew one-third of his total balance. Therefore, the amount withdrawn is

1/3 × 6150 = 6150/3 = 2050 RS

After the withdrawal, his remaining balance in the bank will be the new total minus the amount withdrawn.

His remaining balance in the bank account post-withdrawal will be

6150 RS - 2050 RS = 4100 RS

Response:

First, subtract one-half from each side of the equation.

Next, divide both sides by 6/7.

Then, multiply each side by 7/6.

Detailed explanation:

I simply took the question on ed.

Answer:

Step-by-step explanation:

There are 15 antennas in total.

Out of these, 3 are defective.

This means that 12 antennas are functioning: 15-3=12.

To ensure that no two defective antennas are adjacent, we need to have only one defective at a time placed between the functional ones.

So,

We align the 13 functional antennas, then look for the spaces where the defective antennas can fit

__G __ G __ G __ G __ G __G __ G __ G __ G __ G __ G __ G __G __

Each gap represented by an underscore indicates a possible location for a defective antenna, allowing for just one per space.

Consequently, there are 14 potential spots for the defective antennas. With 3 defectives, we are dealing with a combinatorial arrangement.

ⁿCr= n!/(n-r)!r!

The total number of arrangements possible is

14C3=14!/(14-3)!3!

14C3=14×13×12×11!/11!×3×2

14C3=14×13×12/6

That gives us 364 distinct ways to arrange them.