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2 years ago

15

My tens digit is less than 7. I am one away from being a multiple of 5. If you half me you will not get a whole number. Who am I

? A.883B.429C.133D.524E.635F.236G.712H.671

2 answers:

Hitman42 [59]2 years ago

8 0

Hope this helps, can i get brainliest please

Len [333]2 years ago

6 0

Answer:

its B.429

Step-by-step explanation:

20<70

429+1=multiple of 5

429÷2=214.5

so the answer is 429

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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

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2 years ago

Read 2 more answers