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Olin [163]
2 years ago
11

Hello can u tell me if the answer and work is correct if not tell me

Mathematics
2 answers:
ohaa [14]2 years ago
4 0
Your work seems to be correct
Umnica [9.8K]2 years ago
3 0
Hi.
Your work appears to be correct. If you still doubt it,try running your answers through a calculator, or try plotting the points on a line to see if it corresponds with the other points

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The length of the transverse axis is 6 and the length of the red line segment is 14 how long is the blue line segment
lapo4ka [179]

Answer:

8

Step-by-step explanation:

it is not longer then the red line or equal but is bigger then the green line so there is how i got 8

6 0
3 years ago
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Which is the graph of y – 3 = (x + 6)?
ivann1987 [24]

Answer:

try to use socratic

Step-by-step explanation:

7 0
2 years ago
Which pairs of angles in the figure below are vertical angles?
ch4aika [34]

Answer: Option A and Option C.

Step-by-step explanation:

For this exercise it is important to know the definition of "Vertical Angles".

When two lines intersect or cross, there are a pair of angles that share the same vertex and they are opposite each other. This pair of angles are known as "Vertical angles".

By definition, Vertical angles are congruent, which means that the have the equal measure.

In this case, you can observe in the picture provided in the exercise that the line TI and the line WN intersect each other at the point S.

You can identify that the pair of angles that are opposite to each other and share the same vertex are the shown below:

\angle ISN  and \angle TSW

\angle TSN and \angle ISW

6 0
3 years ago
What is 54x > 87? I need to know fast
RoseWind [281]
X > 29/18

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6 0
3 years ago
a port and a radar station are 2 mi apart on a straight shore running east and west. a ship leaves the port at noon traveling no
Rom4ik [11]

Answer:

The rate of change of the tracking angle is 0.05599 rad/sec

Step-by-step explanation:

Here the ship is traveling at 15 mi/hr north east and

Port to Radar station = 2 miles

Distance traveled by the ship in 30 minutes = 0.5 * 15 = 7.5 miles

Therefore the ship, port and radar makes a triangle with sides

2, 7.5 and x

The value of x is gotten from cosine rule as follows

x² = 2² + 7.5² - 2*2*7.5*cos(45) = 39.04

x = 6.25 miles

By sine rule we have

\frac{sin A}{a} = \frac{sin B}{b}

Therefore,

\frac{sin 45}{6.25} = \frac{sin \alpha }{7.5}

α = Angle between radar and ship α

∴ α = 58.052

Where we put

\frac{sin 45}{6.25} = \frac{sin \alpha }{x} to get

\frac{x}{6.25} = \frac{sin \alpha }{sin 45} and differentiate to get

\frac{\frac{dx}{dt} }{6.25} = cos\alpha\frac{\frac{d\alpha }{dt}  }{sin 45}

\frac{15sin45 }{6.25cos\alpha } =\frac{d\alpha }{dt}  }= 3.208 degrees/second = 0.05599 rad/sec.

6 0
3 years ago
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