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cluponka [151]
3 years ago
13

Solve for m: E=mc² please help

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0

Step-by-step explanation:

we have,

E = mc2

or, E/c2 = m

here is your answer..

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Answer: 15 dimes and 13 nickels

Step-by-step explanation:

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4 years ago
Solve for y 3x-4y=10
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3x-4y=10

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The angle measures associated with which set of ordered pairs share the same reference angle? (Negative StartFraction StartRoot
Katarina [22]

Answer:

(C)\left(-\dfrac{1 }{2},-\dfrac{\sqrt{3} }{2} \right)$ and \left(\dfrac{1 }{2},\dfrac{\sqrt{3} }{2} \right)

Step-by-step explanation:

The reference angle is the angle that the given angle makes with the x-axis.

For an ordered pair to share the same reference angle, the x and y coordinates must be the same or a factor of each other.

From the given options:

(A)\left(-\dfrac{\sqrt{3} }{2} ,-\dfrac{1 }{2}\right)$ and \left(-\dfrac{1 }{2},-\dfrac{\sqrt{3} }{2} \right)\\\\(B)\left(\dfrac{1 }{2},-\dfrac{\sqrt{3} }{2} \right)$ and \left(-\dfrac{\sqrt{3} }{2}, \dfrac{1 }{2}\right)\\\\(C)\left(-\dfrac{1 }{2},-\dfrac{\sqrt{3} }{2} \right)$ and \left(\dfrac{1 }{2},\dfrac{\sqrt{3} }{2} \right)\\\\(D)\left(\dfrac{\sqrt{3} }{2},\dfrac{1 }{2} \right)$ and \left(\dfrac{1 }{2},\dfrac{\sqrt{3} }{2} \right)

We observe that only the pair in option C has the same x and y coordinate with the second set of points being a negative factor of the first term. Therefore, they have the same reference angle.

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3 years ago
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<span>a mathematical expression or function so related to another that their product is one; the quantity obtained by dividing the number one by a given quantity.
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In circle M, segment AB is tangent to the circle at point C. AB has endpoints such that AM BM  ,
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Answer:

6

Step-by-step explanation:

We can use the geometric mean theorem:

The altitude on the hypotenuse is the geometric mean of the two segments it creates.

In your triangle, the altitude is the radius CM and the segments are AC and BC.

CM = \sqrt{AC \times BC} = \sqrt{ 9 \times 4} = \sqrt{36} = \mathbf{6}\\\text{The radius of the circle M is $\large \boxed{\mathbf{6}}$}

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