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Ludmilka [50]
3 years ago
13

Just need help 1-3 because it’s not easy

Mathematics
1 answer:
neonofarm [45]3 years ago
7 0
1 Yes
2 No
3 No
Hope this helps
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I need this one, answer please?
Stolb23 [73]

Answer:

Pair 1.

Step-by-step explanation:

On pair 1, shape A can go into shape B with a rotation.

Pair 2 would be a translation.

Pair 3 would be a reflection across the x-axis.

Pair 4 would be a reflection across the y-axis.

6 0
3 years ago
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But I said not to do that-​
harkovskaia [24]

Answer:

Some people just put those links for virus so don't click on them, and i don't think they actually read the question.

5 0
3 years ago
Point K is on line JL. given JL =4x, JK= 2x+3, and KL=x, determine the numerical length of kl​
balandron [24]

Hi! I'm happy to help!

Our total line is JL (4x), and it is split into two parts: JK, and KL. We have our values, and we know that JK+KL=JL, so we can substitute our values and solve for x:

4x=(2x+3)+(x)

4x=3x+3

To solve for x, we have to isolate it on one side of the equation.

First, let's subtract 3x from both sides so that we can isolate x:

4x=3x+3

-3x -3x

x=3

<u>So, our x=3, which means that KL=3.</u>

I hope this was helpful, keep learning! :D

5 0
3 years ago
Which equation could be used to find the length of the hypotenuse?
Komok [63]

Answer:

<h3>(base)² + (altitude)² = (hypotenuse) ²</h3>

Therefore,

2²+5² = c² will be matched.

7 0
3 years ago
LORAN is a long range hyperbolic navigation system. Suppose two LORAN transmitters are located at the coordinates (−60,0) and (6
USPshnik [31]

LORAN follows an hyperbolic path.

The equation of the hyperbola is: \mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}

The coordinates are given as:

\mathbf{(x,y) = (-60,0)\ (60,0)}

The center of the hyperbola  is  

\mathbf{(h,k) = (0,0)}

The distance from the center to the focal points is given as:

\mathbf{c = 60}

Square both sides

\mathbf{c^2 = 3600}

The distance from the receiver to the transmitters  is given as:

\mathbf{2a = 100}

Divide both sides by 2

\mathbf{a = 50}

Square both sides

\mathbf{a^2 = 2500}

We have:

\mathbf{b^2 = c^2 - a^2}

This gives

\mathbf{b^2 = 3600 - 2500}

\mathbf{b^2 = 1100}  

The equation of an hyperbola is:

\mathbf{\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1}

So, we have:

\mathbf{\frac{(x - 0)^2}{2500} + \frac{(y - 0)^2}{1100} = 1}

\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}

Hence, the equation of the hyperbola is: \mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}

Read more about hyperbolas at:

brainly.com/question/15697124

7 0
3 years ago
Read 2 more answers
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