I need help with these problems. I've gone over the material a lot, and I just don't understand! Please help fast, test is tomorrow!
1. Factor 4m²+40m+100
2. The area of a circle is given by the equation A = ð(25x²+10x+1). Find an expression for the radius.
3. The bed of a pond can be modelled by 30 y= x²-15x+14 where x and y are measured by meters and the x-axis matches the water level of the pond. What is the width of the pond?
4. Find the term to create a perfect square trinomial: x² + 22x
5. Complete the square: r² - 4r - 7 = 0
6. The length of a rectangular swimming pool s x + 30, where x = width. It's area is 2800 square feet. Find the length and the width.
Sure, I'd be happy to help you with these problems. Let's go through each problem step by step and explain how to find the solution.
1. Factor 4m²+40m+100:
To factor this quadratic expression, we can look for two numbers that multiply to give 100 and add up to 40. In this case, the two numbers are 10 and 10. Using these numbers, we can rewrite the expression as (2m+10)(2m+10), which can be further simplified as (2m+10)².
2. The area of a circle is given by the equation A = ð(25x²+10x+1). Find an expression for the radius:
To find the expression for the radius, we need to rearrange the given equation. The formula for the area of a circle is A = ðr², where r is the radius. By equating this formula to the given equation, we have ðr² = ð(25x²+10x+1). Simplifying this equation, we get r² = 25x²+10x+1. Finally, taking the square root of both sides, we get the expression for the radius as r = √(25x²+10x+1).
3. The bed of a pond can be modeled by 30y = x²-15x+14, where x and y are measured in meters and the x-axis matches the water level of the pond. What is the width of the pond?
To find the width of the pond, we need to find the x-intercepts of the equation. The x-intercepts correspond to the points where the bed of the pond intersects the x-axis. To find the x-intercepts, we set y = 0 and solve for x. So we have 30(0) = x²-15x+14. Simplifying, we get x²-15x+14=0. Now we can factor the quadratic expression: (x-1)(x-14) = 0. Setting each factor equal to zero, we get x-1 = 0 and x-14=0. Solving for x, we find x=1 and x=14. Therefore, the width of the pond is 14-1 = 13 meters.
4. Find the term to create a perfect square trinomial: x² + 22x:
To create a perfect square trinomial, we need to take half of the coefficient of the x-term and square it, then add it to the expression. Half of 22 is 11, so we add 11 squared to the expression: x² + 22x + 11² = x² + 22x + 121. Therefore, the term to create a perfect square trinomial is 121.
5. Complete the square: r² - 4r - 7 = 0:
To complete the square for this quadratic equation, we need to add a constant term to both sides of the equation. The constant term is half the coefficient of the linear term squared. In this case, it's (-4/2)² = 4. Adding 4 to both sides of the equation, we get r² - 4r + 4 - 7 + 4 = 0. Simplifying, we have (r-2)² - 3 = 0. Therefore, the equation can be rewritten as (r-2)² = 3.
6. The length of a rectangular swimming pool is x + 30, where x is the width. Its area is 2800 square feet. Find the length and the width:
The area of a rectangle is given by the formula length times width. In this case, the area is given as 2800 square feet, so we have the equation (x+30)(x) = 2800. Expanding and simplifying, we get x² + 30x - 2800 = 0. Now we can solve this quadratic equation by factoring or by using the quadratic formula. Factoring, we find (x+80)(x-50) = 0. Setting each factor equal to zero, we have x+80 = 0 and x-50 = 0. Solving for x, we get x = -80 and x = 50. Since the width cannot be negative, the width of the pool is 50 feet. Substituting this value back into the equation, we can find the length: L = x + 30 = 50 + 30 = 80 feet. Therefore, the length of the pool is 80 feet and the width is 50 feet.
I hope this explanation helps you understand how to solve these problems. Let me know if you have any further questions!