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elena-14-01-66 [18.8K]

2 years ago

9

A plane flying horizontally at an altitude of 3 mi and a speed of 440 mi/h passes directly over a radar station. find the rate a

t which the distance from the plane to the station is increasing when it is 4 mi away from the station. (round your answer to the nearest whole number.)

1 answer:

Licemer1 [7]2 years ago

8 0

The distance is increasing at a rate that is the speed of the plane multiplied by the cosine of the angle between its flight path and the direct line to the radar station. That cosine is 4/√(3²+4²) = 4/5, so the distance is increasing at
440 mi/h × 4/5 = 352 mi/h

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NEED HELP!!!!!!!!!!!!!

Nataliya [291]

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5 0

2 years ago

Express the radical using the imaginary unit, i

Kipish [7]

Answer:

7i.

Step-by-step explanation:

Note : The given expression must be $\sqrt{-49}$ .

We have to write this radical value in the imaginary value.

The given value can be rewritten as

$\sqrt{-49}=\sqrt{49\times (-1)}$

$\sqrt{-49}=\sqrt{49}\times \sqrt{-1}$ $[\because \sqrt{ab}=\sqrt{a}\sqrt{b}]$

$\sqrt{-49}=7\times i$ $[\because \sqrt{-1}=i]$

$\sqrt{-49}=7i$

Therefore, the required expression is 7i.

8 0

2 years ago

Line h contains the points (3,0) and (0,4). which point lies on a line that is perpendicular to line h and passes through point

Iteru [2.4K]

(3,0)(0,4)
slope = (4 - 0) / (0 - 3) = -4/3
A perpendicular line will have a negative reciprocal slope. So our perpendicular line has a slope of 3/4

y = mx + b
slope(m) = 3/4
(-6,-5)....x = -6 and y = -5
now sub into the formula and find b, the y int
-5 = 3/4(-6) + b
-5 = -18/4 + b
-5 + 18/4 = b
-20/4 + 18/4 = b
-2/4 = b
so ur perpendicular line is : y = 3/4x - 2/4....or 3x - 4y = 2

and ur point (6,4) lies on the perpendicular line <===

6 0

2 years ago

HELP ASAP PLEASE WILL MARK BRAINIEST

allsm [11]

It has to be 45 degrees pal

6 0

2 years ago

Simplify the following

den301095 [7]

Answer:

1) 11 $\sqrt{3}$

2) 2 $\sqrt{2}$

3) $20\sqrt{3} + 15\sqrt{2}$

4) $53 + 12\sqrt{10}$

5) -2

6) $7\sqrt{2} - 5\sqrt{3}$

Step-by-step explanation:

1) 2 $\sqrt{12}$ + 3 $\sqrt{48}$ - $\sqrt{75}$

=(2 × 2 $\sqrt{3}$ )+ (3 × 4 $\sqrt{3}$ ) - 5 $\sqrt{3}$

= 4 $\sqrt{3}$ + 12 $\sqrt{3}$ - 5 $\sqrt{3}$

= 11 $\sqrt{3}$

2) 4 $\sqrt{8}$ -2 $\sqrt{98}$ + $\sqrt{128}$

= (4 × 2 $\sqrt{2}$ ) - (2 × 7 $\sqrt{2}$ ) + 8 $\sqrt{2}$

= 8 $\sqrt{2}$ - 14 $\sqrt{2}$ +8 $\sqrt{2}$

= 2 $\sqrt{2}$

3) 5 $\sqrt{12\\}$ - 3 $\sqrt{18} + 4 \sqrt{72} +2\sqrt{75}$

= 5× $2\sqrt{3}$ - 3× $3\sqrt{2}$ + 4× $6\sqrt{2}$ + 2× $5\sqrt{3}$

= $10\sqrt{3} - 9\sqrt{2} +24\sqrt{2} +10\sqrt{3}$

= $20\sqrt{3} + 15\sqrt{2}$

4) $(2\sqrt{2} + 3\sqrt{5} )^{2}$

= $8 + 12\sqrt{10} + 45$

= $53 + 12\sqrt{10}$

5) $(1+\sqrt{3} ) (1-\sqrt{3} )$

= $1 - 3$

= -2

6) $(2\sqrt{6} -1) (\sqrt{3} -\sqrt{2} )$

= $2\sqrt{18}-2\sqrt{12} -\sqrt{3} +\sqrt{2}$

= 2× $3\sqrt{2}$ - 2× $2\sqrt{3}$ - $\sqrt{3} + \sqrt{2}$

= $6\sqrt{2} - 4\sqrt{3} -\sqrt{3} +\sqrt{2}$

= $7\sqrt{2} - 5\sqrt{3}$

Hope the working out is clear and will help you. :)

5 0

2 years ago

Read 2 more answers