Answer:
25
Step-by-step explanation:
When we write expressions for the total cost of each field visit and set them equal, we find the solution to be the ratio of the difference in fixed cost to the difference in variable cost.
y = 75 +7x . . . . . cost for x students to visit the science center
y = 50 +8x . . . . cost for x students to visit the natural history museum
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Subtracting the first equation from the second, we get ...
0 = -25 +x
25 = x . . . . . add 25; the number of students such that costs are equal
The cost will be the same either place for 25 students.
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<em>Additional comment</em>
Here, the fixed cost difference is 75-50=25, and the variable cost difference is 8-7=1. The ratio of these costs is ...
$25/($1 /student) = 25 students.
This relationship only holds when the higher fixed cost is associated with the lower variable cost. Charges are such that one place caters to larger numbers of students (science center), and one prefers fewer students (natural history museum).
100 seconds. The numbers on the left are seconds.
<span>Two triangles with two sides and a non-included angle equal may or may not be congruent. If two angles on one triangle are equal, respectively, to two angles on another triangle, then the triangles are similar, but not necessarily congruent.</span>
<span>x + 9 = 2(x - 1)^2
x + 9 = 2(x^2 - 2x + 1)
x + 9 = 2x^2 - 4x + 2
</span>2x^2 - 4x + 2 - x - 9 = 0
2x^2 - 5x -7 =0
Since the height of an equilateral triangle in terms of its side s is s√3/2, the height of the triangle is 6√3/2 = 3√3 and so the area is (1/2)(6)(3√3) = 9√3.
<span>If we draw a horizontal line a height of h from the base of the triangle, the region is split into two regions: the lower region consisting of a trapezoid of height h and the upper region consisting of a triangle of height 3√3 - h. </span>
<span>Since the upper triangle and the triangle itself are similar triangles, the base and height are proportional. If we let x denote the base of the length of the upper triangle, we have: </span>
<span>(S. of small triangle)/(S. of big triangle) = (Ht. of small triangle)/(Ht. of big triangle) </span>
<span>==> x/6 = (3√3 - h)/(3√3) </span>
<span>==> x = (6√3 - 2h)/√3 </span>
<span>Thus, the area of the upper triangle is: </span>
<span>A = (1/2)[(6√3 - 2h)/√3](3√3 - h) = [(6√3 - 2h)(3√3 - h)]/(2√3). </span>
<span>(Made a dumb mistake about the height here for some reason) </span>
<span>Since we require that the area of this triangle is to be half of the total area (9√3/2), we need to solve: </span>
<span>[(6√3 - 2h)(3√3 - h)]/(2√3) = 9√3/2 </span>
<span>==> (6√3 - 2h)(3√3 - h) = 27 </span>
<span>==> 54 - 6h√3 - 6h√3 + 2h^2 = 27 </span>
<span>==> 2h^2 - 12h√3 + 27 = 0. </span>
<span>Solving with the Quadratic Formula gives: </span>
<span>h = (6√3 + 3√6)/2 ≈ 8.87 units and h = (6√3 - 3√6)/2 ≈ 1.52 units. </span>
<span>Since h = (6√3 + 3√6)/2 would place the line outside of the triangle, we pick h = (6√3 - 3√6)/2. </span>
<span>Therefore, the line should be ==> (6√3 - 3√6)/2 units from the base. </span>
<span>I hope this helps! ^^ Brainliest Please?</span><span>
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