4. lawrence is increasing his rectangular patio in his backyard. his patio is currently 12 feet by 10 feet…

4. lawrence is increasing his rectangular patio in his backyard. his patio is currently 12 feet by 10 feet. he wants to increase the patio by adding a decorative tile the same width (x) all the way around creating a total area of 180 square feet. select all the quadratic equations that represent lawrences new patio area.\n ① (x^{2}+11x - 15 = 0)\n ② (x^{2}-11x - 15 = 0)\n ③ (x^{2}+22x - 60 = 0)\n ④ (x^{2}-22x - 60 = 0)\n ⑤ (4x^{2}+44x - 60 = 0)\n ⑥ (4x^{2}-44x - 60 = 0)\n5. mr. cornells physics class made a catapult to shoot watermelons. a watermelon starting launched was 36 feet above the ground and reached a maximum height of 60.5 feet after 3.5 seconds.\npart a: which of the following equations can represent the height of the watermelon in feet, for x seconds?\n ① (0=-2(x + 3.5)^{2}+60.5)\n ② (0=-2(x - 3.5)^{2}+60.5)\n ③ (0 = 0.5(x - 60.5)^{2}+36)\n ④ (0=-0.5(x - 60.5)^{2}+36)\npart b: after how many second will it take for the watermelon to return to the ground?

4. lawrence is increasing his rectangular patio in his backyard. his patio is currently 12 feet by 10 feet. he wants to increase the patio by adding a decorative tile the same width (x) all the way around creating a total area of 180 square feet. select all the quadratic equations that represent lawrences new patio area.\n ① (x^{2}+11x - 15 = 0)\n ② (x^{2}-11x - 15 = 0)\n ③ (x^{2}+22x - 60 = 0)\n ④ (x^{2}-22x - 60 = 0)\n ⑤ (4x^{2}+44x - 60 = 0)\n ⑥ (4x^{2}-44x - 60 = 0)\n5. mr. cornells physics class made a catapult to shoot watermelons. a watermelon starting launched was 36 feet above the ground and reached a maximum height of 60.5 feet after 3.5 seconds.\npart a: which of the following equations can represent the height of the watermelon in feet, for x seconds?\n ① (0=-2(x + 3.5)^{2}+60.5)\n ② (0=-2(x - 3.5)^{2}+60.5)\n ③ (0 = 0.5(x - 60.5)^{2}+36)\n ④ (0=-0.5(x - 60.5)^{2}+36)\npart b: after how many second will it take for the watermelon to return to the ground?

Answer

Answer:

Question 4:

E. $4x^{2}+44x - 60 = 0$

Question 5 Part A:

B. $0=-2(x - 3.5)^{2}+60.5$

Question 5 Part B:

7

Explanation:

Question 4 Step1: Find new dimensions

The new length is $12 + 2x$ and new width is $10+2x$.

Question 4 Step2: Calculate new - area

Area $A=(12 + 2x)(10 + 2x)=180$.

Question 4 Step3: Expand the equation

$(12 + 2x)(10 + 2x)=120+24x+20x + 4x^{2}=120 + 44x+4x^{2}$. So $4x^{2}+44x+120 = 180$, which simplifies to $4x^{2}+44x - 60 = 0$.

Question 5 Part A Step1: Recall vertex - form of quadratic

The vertex - form of a quadratic is $y=a(x - h)^{2}+k$, where $(h,k)$ is the vertex. The vertex is $(3.5,60.5)$ and since the parabola opens down (maximum height), $a<0$. Substituting into the vertex - form gives $y=-2(x - 3.5)^{2}+60.5$. When $y = 0$ (height above ground), we have $0=-2(x - 3.5)^{2}+60.5$.

Question 5 Part B Step1: Set the equation to zero

We have $0=-2(x - 3.5)^{2}+60.5$. First, move $60.5$ to the other side: $2(x - 3.5)^{2}=60.5$.

Question 5 Part B Step2: Solve for $(x - 3.5)^{2}$

$(x - 3.5)^{2}=\frac{60.5}{2}=30.25$.

Question 5 Part B Step3: Take square - root

$x-3.5=\pm\sqrt{30.25}=\pm5.5$.

Question 5 Part B Step4: Solve for $x$

We get two solutions: $x=3.5 + 5.5=9$ and $x=3.5-5.5=-2$. Since time cannot be negative, the time it takes to return to the ground is $7$ seconds (using the symmetry of the parabola, the time from launch to maximum and from maximum to ground is the same, and the time from launch to maximum is $3.5$ seconds, so total time is $2\times3.5 = 7$ seconds).