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Sergio [31]
3 years ago
10

Which shows the pre-image of triangle X’Y’Z’ before the figure is rotated 90° about the origin

Mathematics
2 answers:
Xelga [282]3 years ago
3 0

Answer:

From the figure shown the coordinates of W'X'Y'Z' are W'(2, 3), X'(6, 3), Y'(6, 4) and Z'(3, 5)

The coordinates of the pre-image that yeilded W'X'Y'Z' according to the rule (-x, -y) can be obtained by changing the signs of the coordinates of W'X'Y'Z'.

Thus, the coordinates of the pre-image that yeilded W'X'Y'Z' are W(-2, -3), X(-6, -3), Y(-6, -4) and Z(-3, -5).

Therefore, the correct graph is graph c.

(Need to update complete question)

sattari [20]3 years ago
3 0

Answer:

C

Step-by-step explanation:

Ed2020

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Answer:what are answer choices

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4 0
3 years ago
0=x2+4x-5 using zero product property
Elenna [48]

Answer:

x=1,-5x=1,−5

Step-by-step explanation:

0=(x−1)(x+5)

2 Switch sides.

(x-1)(x+5)=0(x−1)(x+5)=0

3 Solve for xx.

x=1,-5x=1,−5

4 0
3 years ago
when constructing the incenter of a triangle, The inersection of all three of which type of line needs to be found
lesantik [10]
Answer: Angle bisector

The angle bisector is the line that cuts an angle in half. The three angle bisectors all intersect at the same point which is the incenter. The incenter is the center of the circle in which is largest possible, yet it's fully inside the triangle. No part of this circle is outside the triangle. 
3 0
3 years ago
Decide whether it is possible for a triangle to have the three angle measures or three side lengths given.
9966 [12]

Answer:

Step-by-step explanation:

the first one cant be a triangle

im pretty sure the second one can

the third one can

im pretty sure the fourth one cant

8 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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