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Taya2010 [7]
3 years ago
12

An airplane is flying at an altitude of 40,000 feet. To avoid turbulence the pilot descends at a rate of 2,000 feet / minute for

3 minutes. He continues flying at this altitude until he decides the turbulence is over; then he ascends at a rate of 1,500 feet minute for 5.5 minutes. At what altitude is the plane now flying?​
Mathematics
1 answer:
FromTheMoon [43]3 years ago
5 0

Answer:

The plane is now flying at an altitude of 42,250 feet.

Step-by-step explanation:

Initially, he is at an altitude of 40,000 feet.

Descends at a rate of 2,000 feet / minute for 3 minutes.

So his altitude is now:

40,000 - 3*2,000 = 34,000

Ascends at a rate of 1,500 feet minute for 5.5 minutes.

The plane is now flying at an altitude of:

34,000 + 1,500*5.5 = 42,250

The plane is now flying at an altitude of 42,250 feet.

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Derek has a white cube and a red cube.
cluponka [151]

Answer:

1/72

Step-by-step explanation:

Probability of getting a 4 on white cube is:

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P(R) · P(>4) = 1/2 × 2/6 = 1/6

multiply both probabilities:  1/12 × 1/6 = 1/72

3 0
2 years ago
How do you determine where f(x)=cos^(-1)(lnx) is continuous?
olga nikolaevna [1]
\ln x is continuous over its domain, all real x>0.

Meanwhile, \cos^{-1}y is defined for real -1\le y\le1.

If y=\ln x, then we have -1\le \ln x\le1\implies \dfrac1e\le x\le e as the domain of \cos^{-1}(\ln x).

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7 0
3 years ago
Based on the graph what are the solutions to ax^2 + bx+c=0 Select all that apply
AlexFokin [52]

Answer:

x=-2 and x = 5

Step-by-step explanation:

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8 0
3 years ago
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Write an equation in slope-intercept form for the line that passes through the point  ( -1 , -2 )  and is perpendicular to the l
mamaluj [8]

The equation in slope-intercept form for the line that passes through the point  ( -1 , -2 )  and is perpendicular to the line − 4 x − 3 y  =  − 5 is y = \frac{3}{4}x - \frac{5}{4}

<em><u>Solution:</u></em>

<em><u>The slope intercept form is given as:</u></em>

y = mx + c ----- eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Given that the line that passes through the point  ( -1 , -2 )  and is perpendicular to the line − 4 x − 3 y  =  − 5

Given line is perpendicular to  − 4 x − 3 y  =  − 5

− 4 x − 3 y  =  − 5

-3y = 4x - 5

3y = -4x + 5

y = \frac{-4x}{3} + \frac{5}{3}

On comparing the above equation with eqn 1, we get,

m = \frac{-4}{3}

We know that product of slope of a line and slope of line perpendicular to it is -1

\frac{-4}{3} \times \text{ slope of line perpendicular to it}= -1\\\\\text{ slope of line perpendicular to it} = \frac{3}{4}

Given point is (-1, -2)

Now we have to find the equation of line passing through (-1, -2) with slope m = \frac{3}{4}

Substitute (x, y) = (-1, -2) and m = 3/4 in eqn 1

-2 = \frac{3}{4}(-1) + c\\\\-2 = \frac{-3}{4} + c\\\\c = - 2 + \frac{3}{4}\\\\c = \frac{-5}{4}

\text{ substitute } c = \frac{-5}{4} \text{ and } m = \frac{3}{4} \text{ in eqn 1}

y = \frac{3}{4} \times x + \frac{-5}{4}\\\\y = \frac{3}{4}x - \frac{5}{4}

Thus the required equation of line is found

8 0
3 years ago
Please help ill give you. brainiest
Bond [772]

I believe the answer is b

3 0
3 years ago
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