Three people toss a fair coin and the odd one pays for coffee. if the coins all turn up the same, they are tossed again. find th

Hey there! STEP 1: 1. Simplifying (4x + 3y)² results in (4x - 3y) 1. STEP 2: The equation results in (4x-3y) 1 divided by (16(x²) + 24xy + 9(y²)). STEP 3: Following similar resolutions results in repeating the process of simplification. After completing these steps, we factor and figure that (4x + 3y) is a perfect square, yielding (4x + 3y)². Hence, after numerous simplifications, we achieve that the result concludes to be 4x-3y. I hope this helps!

1) You can depend solely on your braking system, as the vehicle will only travel 250ft from when you apply the brakes until it completely stops, leaving it 50ft away from the cow. 2) Refer to the attached image. j(t) represents the distance from the brake application point after t seconds in feet. j'(t) indicates the car's speed t seconds after braking, expressed in ft/s. j"(t) shows the car's acceleration t seconds post-braking in

. 3) Any time beyond t=5.28 will not accurately reflect the car's path, because at that moment, it will have reached a speed of 0ft/s, and without an external force, the car will remain stationary past that point. 4)

(refer to attached image for the graph). Step-by-step breakdown: 1) Here, we need to determine when the vehicle's speed becomes 0, indicating a complete stop. We achieve that by deriving the position function:

. Setting the first derivative to zero gives us:

95-18t=0, from which we solve for t: -18t=-95, leading to t=5.28s. Now we will calculate the position of the vehicle after 5.28 seconds:

This results in the vehicle stopping 250.69ft after brake application, leaving approximately 50ft between the vehicle and the cow when it halts completely, allowing reliance solely on the brakes. 2) For part 2, I derive the function again to obtain:

j"(t)=-18 and then graph them (see attached image). Here, j(t) illustrates the distance post-brake application t seconds later in feet, j'(t) signifies the velocity t seconds after brakes were applied in ft/s, while j"(t) indicates the car's acceleration post-braking in

. 3) Yes, after t=5.28, the models in part (ii) won't accurately represent the vehicle's trajectory, as at that instance the car's speed would be 0ft/s and if no additional force acts on it, the vehicle won't move beyond that point. 4)

(refer to the attached image for the graph).

Answer:

x=2c/3+1

Step-by-step explanation:

First, subtract 2c from both sides.

−3x=−3−2c

The equation has been transformed into standard form.

−3x=−2c−3

Next, divide both sides by −3.

-3x/-3=-2c-3/-3

Dividing by −3 negates the multiplication by −3.

x=-2c-3/-3

Finally, divide −3−2c by −3.

x=2c/3+1

The adjusted equation , indicating that the graph has been altered 4 units leftward and 6 units downward.

Clarification:

The parent equation is .

The transformed equation is .

Applying the function transformation rules, it's clear that the parent function has been modified into the function .

According to the function transformation rules,

indicates shifting the function b units to the left.

Thus, the transformed function reflects a shift 4 units to the left.

Additionally, from the function transformation rules, we understand that,

indicates shifting the function b units downward.

Consequently, the transformed function mirrors a shift of 6 units downward.

In summary, the adjusted equation , shows the graph shifted 4 units to the left and 6 units downward.

Response:c

Detailed explanation: