If B is the midpoint of AC, that means that the distances from A to B and B to C are the same. The distance from A to B is 8x + 4, and the distance from B to C is 10x - 6, so we can set these two equivalent:
8x + 4 = 10x - 6
Solving for x:
4 = 2x - 6 (subtract 8x from both sides)
10 = 2x (add 6 to both sides)
5 = x (divide both sides by 2)
So, x = 5, which is the second option given.
Answer:
![\sqrt3](https://tex.z-dn.net/?f=%5Csqrt3)
Step-by-step explanation:
To find:
The value of
= ?
Solution:
Kindly consider the equilateral
as attached in the answer area.
Let the side of triangle =
units
Let us draw the perpendicular from vertex A to side BC.
It will divide the side BC in two equal parts.
i.e. BD = DC = ![\frac{a}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%7D%7B2%7D)
Using Pythagorean Theorem in
:
![\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}](https://tex.z-dn.net/?f=%5Ctext%7BHypotenuse%7D%5E%7B2%7D%20%3D%20%5Ctext%7BBase%7D%5E%7B2%7D%20%2B%20%5Ctext%7BPerpendicular%7D%5E%7B2%7D)
Side AD = ![\frac{\sqrt3}{2}a](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt3%7D%7B2%7Da)
Using Trigonometric ratio:
![tan\theta = \dfrac{Perpendicular}{Base}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cdfrac%7BPerpendicular%7D%7BBase%7D)
![tanB = \dfrac{AD}{BD}](https://tex.z-dn.net/?f=tanB%20%3D%20%5Cdfrac%7BAD%7D%7BBD%7D)
Putting the values of AD and BD:
![tan60^\circ=\dfrac{\frac{\sqrt3}{2}a}{\frac{1}{2}a}\\\Rightarrow tan60^\circ = \bold{\sqrt3}](https://tex.z-dn.net/?f=tan60%5E%5Ccirc%3D%5Cdfrac%7B%5Cfrac%7B%5Csqrt3%7D%7B2%7Da%7D%7B%5Cfrac%7B1%7D%7B2%7Da%7D%5C%5C%5CRightarrow%20tan60%5E%5Ccirc%20%3D%20%5Cbold%7B%5Csqrt3%7D)
Answer
The distance from the house to the lake is (2.5 hr)*(48 mi/hr) = 120 miles. The rerurn trip would take (120 mi)/(40 mi/hr) = 3 hours. Distance equals rate times time, so the distance to the lake is 2 1/2 hours times 48 miles per hour.
I hope this helps
We can answer the first part of the question not taking intersecting function into account. The domain of
is all the numbers, x∈(-∞, +∞) and the range is y∈(-∞, 36]. We can observe these results with the help of a graph, as well. Since we are talking about the rainbow, the values above the ground level will make sense. In this case, we will take into account the range as it changes between 0 and 36, included and the domain between -6 and 6. Here (0;36) is the y-intercept and (-6;0) and (6;0) are the x-intercepts of the parabola.
Since in our problem, the linear function that intersects parabola is not given, we have to provide it by ourselves according to the conditions of the problem. It could be any line intersecting parabola in two points. One important point is that the y-intercept has to be no more than 36. Considering these conditions, we can set our linear function to be
. We can observe the points that we included in the table (they have been given with orange dots in the graph and the table is attached below). We can see that the values of the function (values of y) are positive. Indeed, we are discussing the part of the rainbow above the ground level.
The system of equations with linear and quadratic functions has got two solutions and we can observe that result from the graph. The solutions are (-5.823; 2.088) and (5.323; 7.662). The solutions are the intersection points.