answer:I want to find the depth of the pan in inches, so I need to relate the volume of the mixture to the area of the base of the pan. The volume of the mixture is proportional to the amount of the recipe, so if I multiply the recipe by 1.5, I also multiply the volume by 1.5. The original volume of the mixture is the product of the length, width, and depth of the original pan: 9 * 13 * 1 = 117 cubic inches. So the new volume of the mixture is 1.5 * 117 = 175.5 cubic inches. The area of the base of the new pan is the product of the length and width: 10 * 15 = 150 square inches. To find the depth of the new pan, I need to divide the volume by the area: 175.5 / 150 = 1.17 inches. This is the answer, expressed as a decimal to the nearest hundredth.,First, we need to determine the volume of the crispy rice treats mixture. The original mixture fills a 9-inch by 13-inch pan to a depth of one inch, so the volume of the original mixture is: 9 inches * 13 inches * 1 inch = 117 cubic inches Now, we are given a recipe that is 1.5 times the original recipe. So, the volume of the new mixture will be 1.5 times the volume of the original mixture: 1.5 * 117 cubic inches = 175.5 cubic inches Next, we need to determine the depth the new mixture will fill in a 10-inch by 15-inch pan. To do this, we need to find the relationship between the volume of the mixture and the area of the pan: Area of the pan = 10 inches * 15 inches = 150 square inches Depth = Volume / Area Depth = 175.5 cubic inches / 150 square inches = 1.17 inches The pan will be filled to a depth of approximately 1.17 inches., First, we need to determine the volume of the original crispy rice treat mixture. The volume V of a rectangular prism (like our pan) is given by the formula V = lwh, where l is the length, w is the width, and h is the height. For the original recipe, the volume V_orig is: V_orig = l * w * h V_orig = 9 inches * 13 inches * 1 inch

answer:The area of the shaded region is 2 square inches. Here's the reasoning: 1. Since A and B are midpoints of congruent segments overline{XY} and overline{XZ}, respectively, we know that XA = AY = YX/2 and XB = BZ = ZX/2. 2. Also, since altitude overline{XC} bisects overline{YZ}, we have YC = CZ. 3. Now let's consider triangle XYC. Since XA = AY and YC = CZ, this triangle is isosceles with base overline{YC}. 4. Therefore, the altitude from X to overline{YC} also bisects the base, so CY = YA. 5. Thus, triangles XAY and XCY are congruent and have equal areas. 6. Similarly, triangles XBZ and XCZ are congruent and have equal areas. 7. Since triangle XYZ is made up of these four smaller triangles, and its area is given as 8 square inches, each of the smaller triangles must have an area of 2 square inches (since there are four of them). 8. However, the shaded region consists of only two of these smaller triangles, so its area is 2 square inches.,Segments overline{AD} and overline{BE} are drawn perpendicular to overline{YZ}. Segments overline{AB}, overline{AC} and overline{BC} divide triangle XYZ into four congruent triangles. Vertical line segments overline{AD}, overline{XC} and overline{BE} divide each of these in half. Three of the eight small triangles are shaded, or frac{3}{8} of triangle XYZ. The shaded area is frac{3}{8}(8) = 3. asy /* AMC8 2002 #20 Solution */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); draw((5,4)--(5,0), linewidth(0.8)); label(scale(0.8)*"X", (5,4), N); label(scale(0.8)*"Y", (0,0), W); label(scale(0.8)*"Z", (10,0), E); label(scale(0.8)*"A", (2.5,2.2), W); label(scale(0.8)*"B", (7.5,2.2), E); label(scale(0.8)*"C", (5,0), S); label(scale(0.8)*"D", (2.5,0), S); label(scale(0.8)*"E", (7.5,0), S); draw((2.5,0)--(2.5,2)--(7.5,2)--(7.5,0)); draw((2.5,2)--(5,0)--(7.5,2)); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white); /asy,I understand that in order to find the area of the shaded region, we need to first find the area of triangle XYZ, then find the area of the trapezium AYCB, and finally subtract the area of trapezium XYCB from the area of triangle XYZ. The area of triangle XYZ is given to be 8 square inches, so we can proceed with this information. Since triangle XYZ is divided into two equal parts by line AB, we can find the area of each of these smaller triangles by dividing 8 by 2, which gives us 4 square inches per triangle. Now, we can find the area of the trapezium AYCB by adding the areas of the two triangles and the base XY which has a length of 10 inches. The width of the trapezium is half the length of the base, so that is 5 inches. Plugging these values into the trapezoid formula, we can calculate the area of the trapezium to be (4*10 + 0)/2 = 20 square inches. Subtracting the area of trapezium XYCB from the area of the triangle XYZ, we get : 4 + 4 - 20 = -12 We get -12 square inches which doesn’t make a sense because the area can’t be negative and there is a mistake in question. Please check it once..

answer:We can solve this system of equations by adding all three equations together: (3a + b + c) + (a + 3b + c) + (a + b + 3c) = -3 + 9 + 19. This simplifies to 5a + 5b + 5c = 25, or a + b + c = 5. Now we have two equations in three variables: a + b + c = 5, 3a + b + c = -3. Subtracting the second equation from the first gives us 2a = 8, or a = 4. Substituting a = 4 into the first original equation, we get 4 + b + c = 5, or b + c = 1. Now we have two equations in two variables: b + c = 1, 3b + c = 9. Subtracting the first equation from the second gives us 2b = 8, or b = 4. Substituting b = 4 into the first equation, we get 4 + c = 1, or c = -3. Therefore, abc = (4)(4)(-3) = -48.,This problem involves a system of three linear equations with three unknowns. One way to solve such a system is to use elimination or substitution methods. I notice that the coefficients of a, b, c are symmetric in the three equations, so I wonder if there is a clever way to exploit that symmetry. I try adding all three equations together, and I get 5a + 5b + 5c = 25. * so I can divide both sides by 5 and get a + b + c = 5. I wonder what happens if I subtract this equation from each of the original equations. I get 2a = -8, 2b = 4, 2c = 14. So I can divide by 2 and get a = -4, b = 2, c = 7. Now I have the values of a, b, c and I can find their product. abc = (-4)(2)(7) = -56.,To find the value of abc, we can solve the system of linear equations for a, b, and c, and then substitute those values into abc. From the given equations, we can multiply the first equation by 2 to eliminate the c term: 3(2a) + 2b + 2c = -6Rightarrow 6a + 2b + 2c = -6 Then, subtract this new equation from the second and third equations to eliminate the c terms: 1) 2a + 3b + c = 9 - (6a + 2b + 2c) = 9 - 6a - 2b - 2c = 15 - 6a - 2b 2) 2a + b + 3c = 19 - (6a + 2b + 2c) = 19 - 6a - 2b - 2c = 25 - 6a - 2b Now we have two equations with a common variable a: 1) 2a + 3b = 15 - c 2) 2a + b = 25 - 3c Subtract the first equation from the second to eliminate a: (2a + b) - (2a + 3b) = (25 - 3c) - (15 - c) b - 2b = 10 + c b = 10 + c Now substitute b = 10 + c into the first equation: 2a + 3(10 + c) = 15 - c 2a + 30 + 3c = 15 - c 2a + 3c = -15 + c + 30 2a + 3c = 15 Now we have one equation with a common variable c: 2a + 3c = 15 Then, substitute b = 10 + c into the third equation: 2a + (10 + c) + 3

answer:Let's denote the roots as r_1 and r_2. We are given that r_1 = 2a and r_2 = b. Since the roots are distinct, we know that r_1 neq r_2. In the context of quadratic polynomials, the roots represent the values at which the polynomial equals zero. According to Vieta's formulas, the sum of the roots is equal to the opposite of the coefficient of the linear term. To apply Vieta's formula in this case, we notice that the linear term of the polynomial p(x) = x^2 + ax + b is equal to a. Thus, we can write: r_1 + r_2 = -a, substituting r_1=2a and r_2=b gives: 2a + b = -a Adding a to both sides, we get: a+2a + b = 0 Simplifying, we get: 3a + b = 0 The sum of a and b is -3a., Since the polynomial has distinct roots, it can be written as (x-2a)(x-b). Comparing this with x^2+ax+b, we have (x-2a)(x-b) = x^2 - (2a+b)x + 2ab. Comparing coefficients, we get -(2a+b) = a Rightarrow b = -a, and 2ab = b Rightarrow 2a(-a) = b Rightarrow 2a^2 = -b. Substituting b = -a into the equation 2a^2 = b, we obtain 2a^2 = -(-a) Rightarrow 2a^2 = a. Thus, a(2a-1) = 0. If a=0, then b=-a=0, and a+b=0. If a = frac{1}{2}, then b = -a = -frac{1}{2}, and a+b = frac{1}{2} - frac{1}{2} = 0. Therefore, in all cases, a+b=0.,I notice that the roots of the polynomial are related to its coefficients in a special way. I wonder if I can use this relationship to find a simple equation involving a and b. I recall that the sum of the roots of a quadratic equation is equal to the negative coefficient of x, and the product of the roots is equal to the constant term. So, in this case, I have 2a+b = -a and 2a cdot b = b. I can simplify the first equation by adding a to both sides, and get 3a+b = 0. I can also simplify the second equation by dividing both sides by b, and get 2a = 1, as long as b neq 0. If b = 0, then the polynomial has a double root at x = 0, which contradicts the assumption that the roots are distinct. So, I can assume that b neq 0 and solve for a. I get a = frac{1}{2}. Now, I can plug this value of a into the equation 3a+b = 0 and solve for b. I get b = -frac{3}{2}. Therefore, the polynomial is p(x) = x^2 + frac{1}{2}x - frac{3}{2}. The question asks for a+b. I can add frac{1}{2} and -frac{3}{2} and get a+b = -1.