The table for the complete ratios between distance and time is shown below.
<h3>What is the Ratio of Distance to Time?</h3>
The ratio of distance to time, given that 280 miles will be covered in 4 hours is: 280/4 = 70 miles per hour.
Thus, for 2 hours: 2(70) = 140 miles distance.
For 350 miles, we would have: 350/70 = 5 hours time.
For 6 hours: 6(70) = 420 miles distance.
For 8 hours: 8(70) = 560 miles distance.
For 700 miles, we would have: 700/70 = 7 hours time.
The complete table is shown in the image below.
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Answer:
0.081
Step-by-step explanation:
Given parameters:
Vertical drop = 70ft
Horizontal distance = 866ft
Unknown:
Slope of the rapids =?
Solution:
Slope is the rise or vertical height divide by the horizonal distance.
Slope = 
Insert the parameters and solve;
Slope = 
= 0.081
<h3>
Answer: Choice F) 1</h3>
=============================================
Explanation:
x^2+8x+7 factors to (x+1)(x+7)
x^2+4x-21 factors to (x-3)(x+7)
Therefore, the numerator (x^2+8x+7)(x-3) updates to (x+1)(x+7)(x-3)
The denominator (x^2+4x-21)(x+1) updates to (x-3)(x+7)(x+1)
We have three pairs of terms that will cancel between the numerator and denominator. The (x+1) terms pair up and cancel out, so do the (x+7) terms and also the (x-3) terms as well. Everything cancels out. The only thing left is 1 because x/x = 1 where x is nonzero. We can have a more complicated expression replace x to get the same basic idea. This is what I mean by "canceling".
Answer:
Sine Formula or Sine Rule :
BC / sinA = AC / sinB
60 / sinA = 84.95 / sin(115°)
60 / sinA = 84.95 / 0.91
60 / sinA = 93.7
sinA = 60 / 93.7
sinA = 0.64
A = arc(sin0.64)
A = 39°48'30"
A ≈ 40°
Hence, the value of angle BAC is <u>40°</u>
- <u>40°</u> is the right answer.
Start with #47. To find the critical values, you must differentiate this function. x times (4-x)^3 is a product, so use the product rule. The derivative comes out to f '(x) = x*3*(4-x)^2*(-1) + (4-x)^3*1 = (4-x)^2 [-3x + 4-x]
Factoring this, f '(x) = (4-x)^2 [-3x+4-x]
Set this derivative equal to zero (0) and solve for the "critical values," which are the roots of f '(x) = (4-x)^2 [-3x+4-x]. (4-x)^2=0 produces the "cv" x=4.
[-3x+ (4-x)] = 0 produces the "cv" x=1. Thus, the "cv" are {4,1}.
Evaluate the given function at x: {4,1}. For example, if x=1, f(1)=(1)(4-1)^3, or 2^3, or 8. Thus, one of the extreme values is (1,8).
