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
Learn more about minutes of distance apart here
brainly.com/question/8783264
7(x+2)
Let's find the common term in this expression to factor it out. it is 7.
Factoring is the opposite of distributing.
If we distribute, we will get the original expression.
Next:
45-27k
Here, the common term is 9:
9(5-3k)
12ab+7b - the common term is b:
b(12a+7)
y^2-9y.
Here the common term is y:
y(y-9)
8r-32r² the common term here is 8r:
8r(1-4r)
16gh+28gh: the Common Term is 4gh:
4gh(4+7)
21w²-77wx The common term is 7w:
7w(3w-11x)
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Answer:
slope = -5
Step-by-step explanation:
Answer:
Hope this helps
Step-by-step explanation:
t=6
Answer:
A cross section parallel to the base is a square measuring 4 cm by 4 cm.
A cross section perpendicular to the base through the midpoints of opposite sides is a square measuring 4 cm by 4 cm.
A cross section that passes through the entire bottom front edge and the entire top back edge is a rectangle measuring 4 cm by greater than 4 cm.
Step-by-step explanation:
Cross sections parallel and Perpendicular to the base are squares 4 × 4
The diagonal cross section will be a rectangle with base 4 and height
sqrt(4² + 4²) = 4sqrt(2) > 4
