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zhuklara [117]
3 years ago
12

A commercial aircraft gets the best fuel efficiency if it operates at a minimum altitude of 29,000 feet and a maximum altitude o

f 41,000 feet. Model the most fuel-efficient altitudes using a compound inequality.
x ≥ 29,000 and x ≤ 41,000
x ≤ 29,000 and x ≥ 41,000
x ≥ 41,000 and x ≥ 29,000
x ≤ 41,000 and x ≤ 29,000
Mathematics
1 answer:
natita [175]3 years ago
5 0

Answer:

x\geq 29,000  and  x\leq 41,000

Step-by-step explanation:

Let

x -----> the altitude of a commercial aircraft

we know that

The expression " A minimum altitude of 29,000 feet" is equal to

x\geq 29,000

All real numbers greater than or equal to 29,000 ft

The expression " A maximum altitude of 41,000 feet" is equal to

x\leq 41,000

All real numbers less than or equal to 41,000 ft

therefore

The compound inequality is equal to

x\geq 29,000  and  x\leq 41,000

All real numbers greater than or equal to 29,000 ft and less than or equal to 41,000 ft

The solution is the interval ------> [29,000,41,000]

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1. Find the domain of the given function. (1 point)
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the answer is C) x ≥ -3, x ≠ 2  
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3. Perform the requested operation or operations.
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 4. Perform the requested operation or operations.
f(x) = x minus five divided by eight. ; g(x) = 8x + 5, find g(f(x)).
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6 0
3 years ago
Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.
Leni [432]

Answer:

a)

g(2.9) \approx -6.6

g(3.1) \approx -3.4

b)

The values are too small since g'' is positive for both values of x in. I'm speaking of the x values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

y-y_1=m(x-x_1)

where m is the slope and (x_1,y_1) is a point on that line.

We want to find the equation of the tangent line of the curve g at the point (3,-5) on g.

So we know (x_1,y_1)=(3,-5).

To find m, we must calculate the derivative of g at x=3:

m=g'(3)=(3)^2+7=9+7=16.

So the equation of the tangent line to curve g at (3,-5) is:

y-(-5)=16(x-3).

I'm going to solve this for y.

y-(-5)=16(x-3)

y+5=16(x-3)

Subtract 5 on both sides:

y=16(x-3)-5

What this means is for values x near x=3 is that:

g(x) \approx 16(x-3)-5.

Let's evaluate this approximation function for g(2.9).

g(2.9) \approx 16(2.9-3)-5

g(2.9) \approx 16(-.1)-5

g(2.9) \approx -1.6-5

g(2.9) \approx -6.6

Let's evaluate this approximation function for g(3.1).

g(3.1) \approx 16(3.1-3)-5

g(3.1) \approx 16(.1)-5

g(3.1) \approx 1.6-5

g(3.1) \approx -3.4

b) To determine if these are over approximations or under approximations I will require the second derivative.

If g'' is positive, then it leads to underestimation (since the curve is concave up at that number).

If g'' is negative, then it leads to overestimation (since the curve is concave down at that number).

g'(x)=x^2+7

g''(x)=2x+0

g''(x)=2x

2x is positive for x>0.

2x is negative for x.

That is, g''(2.9)>0 \text{ and } g''(3.1)>0.

So 2x is positive for both values of x which means that the values we found in part (a) are underestimations.

6 0
3 years ago
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