The number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
- Since one student runs at a speed of 180 feet per minutes, in time t minutes, he moves a distance of d = 180t.
- Also, the second student runs at a speed of 160 feet per minute, in time t minutes, he moves a distance of d' = 160t.
Since they are initially 10 feet apart, their total distance apart after t minutes is D = d + 10 + d'
D = 180t + 10 + 160t
D = 340t + 10
<h3>Number of minutes before they are 1870 feet</h3>
Making t subject of the formula, we have
t = (D - 10)/340
Since they are 1870 feet apart after t minutes, D = 1870 feet.
t = (D - 10)/340
t = (1870 - 10)/340
t = 1860/340
t = 5.47 minutes
So, the number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
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brainly.com/question/8783264
Answer:
Step-by-step explanation:
Since we know we only have 640 feet of fence available, we know that L + W + L = 640, 2L + W = 640. This allows us to represent the width, W, in terms of L: W = 640 – 2L
Remember, the area of a rectangle is equal to the product of its width and length, therefore,
Notice that, quadratic has been vertically reflected, since the coefficient on the squared term is negative, so the graph will open downwards, and the vertex will be a maximum value for the area.
recall,
Since our function is A(L)=640L-2L², we get
plug in the value of a and b into the first formula:
hence,the dimensions of the pen that will maximize the area are<u> </u><u>L=</u><u>1</u><u>6</u><u>0</u><u>m</u><u> </u>and<u> </u><u>W=</u><u>7</u><u>6</u><u>8</u><u>0</u><u>0</u><u>/</u><u>1</u><u>6</u><u>0</u><u>=</u><u>4</u><u>8</u><u>0</u><u>m</u>
and we're done!
Youll have 80 pennies. dime=.10 10x8=80
Answer:
which agrees with option"B" of the possible answers listed
Step-by-step explanation:
Notice that in order to solve this problem (find angle JLF) , we need to find the value of the angle defined by JLG and subtract it from , since they are supplementary angles. So we focus on such, and start by drawing the radii that connects the center of the circle (point "O") to points G and H, in order to observe the central angles that are given to us as and . (see attached image)
We put our efforts into solving the right angle triangle denoted with green borders.
Notice as well, that the triangle JOH that is formed with the two radii and the segment that joins point J to point G, is an isosceles triangle, and therefore the two angles opposite to these equal radius sides, must be equal. We see that angle JOH can be calculated by :
Therefore, the two equal acute angles in the triangle JOH should add to:
resulting then in each small acute angle of measure .
Now referring to the green sided right angle triangle we can find find angle JLG, using:
Finally, the requested measure of angle JLF is obtained via: