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3 years ago

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

Read 2 more answers

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.

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