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kotegsom [21]
3 years ago
7

select all the names appropriate for the angle indicated by the arc mark

Mathematics
1 answer:
gulaghasi [49]3 years ago
8 0

Answer:

The answers are

angle KHM and

angle MHK

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Sketch the situation if necessary and use related rates to solve. Two airplanes are flying in the air at the same height. Airpla
Alexeev081 [22]

Answer:

  -390 mph

Step-by-step explanation:

Let a and b represent, respectively, the distances of A and B from the airport. The distance d between the planes is then given by the Pythagorean theorem as ...

  d² = a² + b²

Differentiating with respect to time, we have ...

  2d·d' = 2a·a' +2b·b'

Solving for d', we get ...

  d' = (a/d)a' +(b/d)b'

The value of d at the time of interest is ...

  d = √(a² +b²) = √(30² +40²) = √2500 = 50

Then the rate of change of separation is ...

  d' = (30/50)(-250 mph) +(40/50)(-300 mph) = (-150 -240) mph

  d' = -390 mph

The distance between planes is decreasing at 390 miles per hour.

4 0
3 years ago
Use the drop-down menus to complete each equation so the statement about its solution is true.
tino4ka555 [31]

Answer:

No solutions: 7−5+3x−1= 3x + 7

One solution: 7−5+3x−1= 2x + 4

Infinitely Many Solutions  7−5+3x−1= 3x + 1

Step-by-step explanation:

No solutions means it results in no x and a false statement like 3 = 0.

One solution means it results in x = a number.

Infinite solutions means its results in no x and in a true statement like 3=3.

  • So to get No solutions: 7−5+3x−1= 3x + 7

This will eliminate the x and result in 1 = 7 which is false.

  • One solution: 7−5+3x−1= 2x + 4

This will result in x= 3

  • Infinitely Many Solutions  7−5+3x−1= 3x + 1

This will eliminate the x and result in 1 = 1 which is true.

7 0
4 years ago
The box plots below show attendance at a local movie theater and high school basketball games: Two box plots are shown. The top
nignag [31]

Answer:

Inter-quartile range

Step-by-step explanation:

When you have the quartile, you can use the IQR (interquartile range) to measure the dispersion.

IQR = Q3 - Q1

Movies: IQR = 195 - 162 = 33

Basketball: IQR = 225 - 170 = 55

Attendance at Basketball is more dispersed/spread out as compared to attendance at Movies

8 0
3 years ago
Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

7 0
3 years ago
What is the equivalent of 1 liter in gallons?
Luba_88 [7]
0.2642 gallons is your amazing answer<span />
4 0
3 years ago
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