The problem involves a geometric configuration with lines and curves, and we are tasked with finding the value of \( a + d \). Let's break down the solution in detail.
The given equation is: \[ x + y + 4 = 0 \] This represents a line with slope \(-1\) and y-intercept \(-4\). The points of intersection of the line with the axes are marked in the figure. - The line intersects the x-axis at \( (-4, 0) \). - The line intersects the y-axis at \( (0, -4) \). The area in question is the area of a right triangle formed by the x-axis, y-axis, and the line \( x + y + 4 = 0 \).
The area of a triangle is given by the formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] In this case, the base of the triangle is the distance from the origin \( O \) to \( A(-4, 0) \), which is 4 units. The height is the distance from the origin \( O \) to the point \( B(0, -4) \), which is also 4 units. Therefore, the area is: \[ \text{Area} = \frac{1}{2} \times 4 \times 5 = 10 \] Hence, we can set \( a = 10 \).
The image shows the relationship between the variables \( a \) and \( d \). Given the geometric configuration, we know that: \[ 6 = 4 \quad \text{so} \quad a + d = 14 \] Thus, the value of \( a + d \) is 14.
\[ a + d = 14 \]
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
Let the focal chord PQ of the parabola $ y^2 = 4x $ make an angle of $ 60^\circ $ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, $ S $ being the focus of the parabola, touches the y-axis at the point $ (0, \alpha) $, then $ 5\alpha^2 $ is equal to:
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).